Sequences and Series Reviewer

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### Sequences and Series This chapter is all about understanding what it means to sum an infinite collection of numbers. We'll explore when an infinite sum (called an infinite series) is meaningful and how to calculate it using algebra and calculus. Infinite series are super important in science and math because many functions naturally appear as infinite series or can be represented by them, which is useful for numerical calculations. Throughout this discussion, we'll assume $N = \{1, 2, 3, 4, \dots\}$. ### Sequences #### Definition 2.1.1. A **sequence function** is a function from $N$ to $R$. An **(infinite) sequence** is the ordered list of elements from the range of a sequence function. The elements are called its **terms**. #### Notation If $f: N \to R$ is a sequence function and $a_n = f(n)$, the resulting sequence is denoted as $\{a_n\}_{n=1}^{+\infty} = \{a_1, a_2, \dots, a_n, \dots\}$ or simply $\{a_n\}$ if there's no confusion. **Variation:** For any $k \in N \cup \{0\}$, $\{a_n\}_{n=k}^{+\infty} = \{a_k, a_{k+1}, a_{k+2}, \dots, a_n, \dots\}$. This means a sequence can start at $n=0$ or any $n > 1$. #### Remark 2.1.2. Below are some special sequences. 1. **Arithmetic sequence:** $\{t_n\}$ where $t_n = a + dn$ for constants $a$ and $d$, with $d \neq 0$. - Example: $\{-2+3n\}_{n=0}^{+\infty} = \{-2, 1, 4, 7, 10, 13, \dots\}$ - Example: $\{3-2n\}_{n=1}^{+\infty} = \{3, 1, -1, -3, -5, \dots\}$ 2. **Geometric sequence:** $\{t_n\}$ where $t_n = ar^n$ for non-zero constants $a$ and $r$. - Example: $\{3 \cdot 2^n\}_{n=0}^{+\infty} = \{3, 6, 12, 24, \dots\}$ - Example: $\{24(\frac{1}{2})^n\}_{n=0}^{+\infty} = \{24, 12, 6, 3, \dots\}$ 3. **Fibonacci sequence:** $\{t_n\}_{n=1}^{+\infty}$ where $t_1 = 1$, $t_2 = 1$, and $t_n = t_{n-1} + t_{n-2}$ for all $n \geq 3$. - Example: $\{1, 1, 2, 3, 5, 8, 13, 21, \dots\}$ Our main goal is to study how a sequence $\{a_n\}$ behaves as $n$ gets really large (increases without bound). #### Illustration 2.1.3. Let's look at the sequence $\{a_n\} = \{\frac{1}{n^2}\}_{n=1}^{+\infty}$. Notice how the values of $a_n$ get closer and closer to $0$ as $n$ increases without bound. Intuitively, we say $a_n$ **converges** to $0$, and that the sequence is **convergent**. | n | $a_n$ | n | $a_n$ | | :--- | :--------- | :------- | :--------------- | | 1 | 1 | 10000 | 0.00000001 | | 10 | 0.01 | 100000 | 0.0000000001 | | 100 | 0.0001 | 1000000 | 0.000000000001 | | 1000 | 0.000001 | 10000000 | 0.00000000000001 | #### Definition 2.1.4. Let $\{a_n\}$ be a sequence and let $L \in R$. The **limit** of $\{a_n\}$ is $L$ if for every $\epsilon > 0$, there exists an $N > 0$ such that $|a_n - L| N$. In this situation, we say the sequence $\{a_n\}$ **converges to $L$** or is **convergent**, and we write $\lim_{n \to +\infty} a_n = L$. If no such number $L$ exists, we say that $\{a_n\}$ **diverges** or is **divergent**. #### Example 2.1.5. Prove using the definition that the sequence $\{\frac{1}{n^2}\}_{n=1}^{+\infty}$ converges to $0$. **Proof:** For any $\epsilon > 0$, let $N = \frac{1}{\sqrt{\epsilon}}$. If $n \in N$ with $n > N$, then $n^2 > \frac{1}{\epsilon}$, so $\frac{1}{n^2} N$. It follows from Definition 2.1.4 that $\{\frac{1}{n^2}\}_{n=1}^{+\infty}$ is convergent with $\lim_{n \to +\infty} \frac{1}{n^2} = 0$. #### Remark 2.1.6. Let $m, k \in N$. The sequence $\{a_n\}_{n=k}^{+\infty}$ converges to $L$ if and only if the sequence $\{a_n\}_{n=m}^{+\infty}$ converges to $L$. #### Theorem 2.1.7. Let $f$ be a function defined on $[k, +\infty)$ for some $k \in N$ and let $a_n = f(n)$ for all $n \in N$ with $n \geq k$. 1. If $\lim_{x \to +\infty} f(x)$ exists, then $\lim_{n \to +\infty} a_n = \lim_{x \to +\infty} f(x)$. 2. If $\lim_{x \to +\infty} f(x) = +\infty$ (respectively, $-\infty$), then $\lim_{n \to +\infty} a_n = +\infty$ (respectively, $-\infty$). #### Example 2.1.8. Prove using Theorem 2.1.7 that the sequence $\{\frac{1}{n^2}\}_{n=1}^{+\infty}$ converges to $0$. **Proof:** The function $f$ given by $f(x) = \frac{1}{x^2}$ is defined on $[1, +\infty)$ and $\lim_{x \to +\infty} f(x) = 0$. It follows from Theorem 2.1.7 that $\{\frac{1}{n^2}\}_{n=1}^{+\infty}$ converges to $0$. #### Example 2.1.9. Determine whether the sequence $\{\frac{\sqrt{n^2-5}}{3n+1}\}_{n=3}^{+\infty}$ is convergent or divergent. **Solution:** Let $f(x) = \frac{\sqrt{x^2-5}}{3x+1}$, which is defined on $[3, +\infty)$. By Theorem 2.1.7, $\lim_{x \to +\infty} \frac{\sqrt{x^2-5}}{3x+1} = \lim_{x \to +\infty} \frac{x\sqrt{1-\frac{5}{x^2}}}{x(3+\frac{1}{x})} = \lim_{x \to +\infty} \frac{\sqrt{1-\frac{5}{x^2}}}{3+\frac{1}{x}} = \frac{1}{3}$. So $\{\frac{\sqrt{n^2-5}}{3n+1}\}_{n=3}^{+\infty}$ is convergent with limit $\frac{1}{3}$. #### Example 2.1.10. Determine whether the sequence $\{\frac{e^n}{n}\}_{n=1}^{+\infty}$ is convergent or divergent. **Solution:** Let $f(x) = \frac{e^x}{x}$, which is continuous on $[1, +\infty)$. Since $f(x)$ is continuous on $[1, +\infty)$ and $\lim_{x \to +\infty} f(x)$ is indeterminate of the form $\frac{\infty}{\infty}$, L'Hopital's Rule can be applied. By Theorem 2.1.7 we have $\lim_{n \to +\infty} \frac{e^n}{n} = \lim_{x \to +\infty} \frac{e^x}{x} \overset{L'HR}{=} \lim_{x \to +\infty} \frac{D_x e^x}{D_x x} = \lim_{x \to +\infty} \frac{e^x}{1} = +\infty$. So the sequence $\{\frac{e^n}{n}\}_{n=1}^{+\infty}$ is divergent. #### Theorem 2.1.11 (Squeeze Theorem for Sequences). Let $\{a_n\}$, $\{b_n\}$, and $\{c_n\}$ be sequences. If $a_n \leq b_n \leq c_n$ for all $n \geq k$ for some $k \in N$, and $\lim_{n \to +\infty} a_n = \lim_{n \to +\infty} c_n = L$ for some $L \in R$, then $\lim_{n \to +\infty} b_n = L$. #### Example 2.1.12. Determine whether the sequence $\{\frac{n!}{n^n}\}_{n=1}^{+\infty}$ is convergent or divergent. **Solution:** Note that the expression $\frac{x!}{x^x}$ is not defined for non-integer values of $x$, so Theorem 2.1.7 cannot be used. We'll use the Squeeze Theorem. Let $a_n = \frac{n!}{n^n}$. For each $n \in N$, the term $a_n$ is positive, and $a_n = \frac{1 \cdot 2 \cdot 3 \cdot \dots \cdot n}{n \cdot n \cdot n \cdot \dots \cdot n} = \frac{1}{n} \cdot \frac{2}{n} \cdot \frac{3}{n} \cdot \dots \cdot \frac{n}{n} \leq \frac{1}{n} \cdot 1 \cdot 1 \cdot \dots \cdot 1 = \frac{1}{n}$. So $0 0$ for all $n$. We have $\frac{a_{n+1}}{a_n} = \frac{2^{n+1}}{(n+1)!} \cdot \frac{n!}{2^n} = \frac{2^{n+1}}{2^n} \cdot \frac{n!}{(n+1)!} = \frac{2}{n+1}$. Since $\frac{2}{n+1} 0$, and $a_n = \frac{1 \cdot 2 \cdot 3 \cdot \dots \cdot n}{n \cdot n \cdot n \cdot \dots \cdot n} = \frac{1}{n} \cdot \frac{2}{n} \cdot \frac{3}{n} \cdot \dots \cdot \frac{n}{n} \leq \frac{1}{n} \cdot 1 \cdot 1 \cdot \dots \cdot 1 = \frac{1}{n}$. So $0 1$. b. Use the Monotone Convergence Theorem to show that the sequence converges when $0 ### Series of Constant Terms #### Definition 2.2.1. An **infinite series** is an expression of the form $\sum_{n=1}^{+\infty} a_n = a_1 + a_2 + \dots + a_n + \dots$. The numbers $a_1, a_2, a_3, \dots$ are called **terms** of the series. #### Notation If there is no ambiguity, we may write $\sum_{n=1}^{+\infty} a_n$ as $\sum a_n$ or simply $\sum a_n$. A **partial sum** of an infinite series $\sum_{n=1}^{+\infty} a_n$ is a finite sum $S_k$ given by $S_k = \sum_{n=1}^k a_n = a_1 + a_2 + \dots + a_k$. #### Definition 2.2.2. Let $S_k$ denote the $k^{th}$ partial sum of a series $\sum_{n=1}^{+\infty} a_n$. The series is said to be: 1. **convergent** with sum $S$ if and only if the sequence $\{S_k\}_{k=1}^{+\infty}$ of partial sums converges to $S$. In this case, we write $\sum_{n=1}^{+\infty} a_n = S$; 2. **divergent** if and only if the sequence of partial sums diverges. #### Example 2.2.3. Let $c \in R$ and consider the constant series $\sum_{n=1}^{+\infty} c$. Then for each $k \in N$, $S_k = \sum_{n=1}^k c = ck$. If $c=0$, then $S_k=0$ and $\lim_{k \to +\infty} S_k = 0$. If $c \neq 0$, then $\lim_{k \to +\infty} S_k = \lim_{k \to +\infty} ck = \begin{cases} +\infty & \text{if } c > 0 \\ -\infty & \text{if } c 1$. Then $\lim_{k \to +\infty} r^{k+1}$ does not exist. It follows that neither does $\lim_{k \to +\infty} S_k$. Hence the series diverges. Thus, the geometric series $\sum_{n=0}^{+\infty} ar^n$ converges with sum $\frac{a}{1-r}$ if $|r| 1 + \int_2^k \frac{1}{x} dx + \dots + \int_{k}^{k+1} \frac{1}{x} dx = 1 + \int_1^{k+1} \frac{1}{x} dx = 1 + [\ln x]_1^{k+1} = 1 + \ln(k+1)$. Since $\lim_{k \to +\infty} \ln(k+1) = +\infty$, we also have $\lim_{k \to +\infty} S_k = +\infty$. So the harmonic series diverges. #### Theorem 2.2.8. 1. Let $c \in R \setminus \{0\}$. If $\sum_{n=1}^{+\infty} a_n$ converges then $\sum_{n=1}^{+\infty} ca_n$ converges with sum $c \sum_{n=1}^{+\infty} a_n$, and if $\sum_{n=1}^{+\infty} a_n$ diverges then $\sum_{n=1}^{+\infty} ca_n$ diverges. 2. If $\sum_{n=1}^{+\infty} a_n$ and $\sum_{n=1}^{+\infty} b_n$ both converge, then $\sum_{n=1}^{+\infty} (a_n + b_n)$ converges with sum $\sum_{n=1}^{+\infty} a_n + \sum_{n=1}^{+\infty} b_n$. 3. If $\sum_{n=1}^{+\infty} a_n$ converges and $\sum_{n=1}^{+\infty} b_n$ diverges, then $\sum_{n=1}^{+\infty} (a_n + b_n)$ diverges. 4. Suppose that $a_n = b_n$ for all but a finite number of values of $n$. If $\sum_{n=1}^{+\infty} a_n$ converges then $\sum_{n=1}^{+\infty} b_n$ converges, and if $\sum_{n=1}^{+\infty} a_n$ diverges then $\sum_{n=1}^{+\infty} b_n$ diverges. #### Example 2.2.9. Determine whether the series $\sum_{n=1}^{+\infty} \frac{2}{n+8}$ converges or diverges. **Solution:** By changing the index of summation, we can write $\sum_{n=1}^{+\infty} \frac{2}{n+8} = \sum_{n=9}^{+\infty} \frac{2}{n} = 2 \sum_{n=9}^{+\infty} \frac{1}{n}$. The series $\sum_{n=9}^{+\infty} \frac{1}{n}$ differs from the harmonic series by exactly eight terms, and so is divergent by statement 4 of Theorem 2.2.8. Since $2 \neq 0$, statement 1 of the theorem implies that $\sum_{n=1}^{+\infty} \frac{2}{n+8}$ is also divergent. #### Example 2.2.10. Determine whether the series $\sum_{n=1}^{+\infty} \frac{n-2^n}{n \cdot 2^n}$ converges or diverges. **Solution:** Notice that $\sum_{n=1}^{+\infty} \frac{n-2^n}{n \cdot 2^n} = \sum_{n=1}^{+\infty} (\frac{n}{n \cdot 2^n} - \frac{2^n}{n \cdot 2^n}) = \sum_{n=1}^{+\infty} (\frac{1}{2^n} - \frac{1}{n})$. The series $\sum_{n=1}^{+\infty} \frac{1}{2^n}$ is geometric with common ratio $\frac{1}{2}$, which has absolute value less than $1$, and hence is convergent. On the other hand, $\sum_{n=1}^{+\infty} \frac{1}{n}$ is the harmonic series, which is divergent. By statement 3 of Theorem 2.2.8, the series $\sum_{n=1}^{+\infty} (\frac{1}{2^n} - \frac{1}{n})$ diverges. #### Theorem 2.2.11 (Divergence Test). If the series $\sum_{n=1}^{+\infty} a_n$ is convergent then $\lim_{n \to +\infty} a_n = 0$. Equivalently, if $\lim_{n \to +\infty} a_n \neq 0$ then $\sum_{n=1}^{+\infty} a_n$ is divergent. **Proof:** Let $S_n = \sum_{k=1}^n a_k$. For $n \geq 2$, $a_n = S_n - S_{n-1}$. If $\sum_{k=1}^{+\infty} a_k$ is convergent then there is a number $S$ such that $\lim_{n \to +\infty} S_n = S$. Therefore $\lim_{n \to +\infty} a_n = \lim_{n \to +\infty} (S_n - S_{n-1}) = \lim_{n \to +\infty} S_n - \lim_{n \to +\infty} S_{n-1} = S - S = 0$. #### Example 2.2.12. Determine whether the series $\sum_{n=1}^{+\infty} \frac{n^2+n-1}{3n^2-2n}$ is convergent or divergent. **Solution:** Since $\lim_{n \to +\infty} \frac{n^2+n-1}{3n^2-2n} = \lim_{n \to +\infty} \frac{n^2(1+\frac{1}{n}-\frac{1}{n^2})}{n^2(3-\frac{2}{n})} = \lim_{n \to +\infty} \frac{1+\frac{1}{n}-\frac{1}{n^2}}{3-\frac{2}{n}} = \frac{1}{3} \neq 0$, the series diverges by the Divergence Test. #### Example 2.2.13. Determine if the series $\sum_{n=1}^{+\infty} \frac{n}{1+\ln(n)}$ converges or diverges. **Solution:** Using L'Hopital's Rule, we get $\lim_{n \to +\infty} \frac{n}{1+\ln(n)} = \lim_{x \to +\infty} \frac{x}{1+\ln(x)} \overset{L'HR}{=} \lim_{x \to +\infty} \frac{1}{\frac{1}{x}} = \lim_{x \to +\infty} x = +\infty$. So the series diverges by the Divergence Test. **Warning:** If $\lim_{n \to +\infty} a_n = 0$, the Divergence Test is inconclusive. #### Example 2.2.14. Consider the geometric series $\sum_{n=0}^{+\infty} (\frac{1}{3})^n$ and the harmonic series $\sum_{n=1}^{+\infty} \frac{1}{n}$. Notice that $\lim_{n \to +\infty} (\frac{1}{3})^n = 0$ and $\lim_{n \to +\infty} \frac{1}{n} = 0$. However, the geometric series converges while the harmonic series diverges. #### Exercises 1. Determine whether the series converges or diverges. If the series converges, find its sum. a. $\sum_{n=0}^{+\infty} (\frac{2}{3})^n$ b. $\sum_{n=0}^{+\infty} (\frac{9}{8})^n$ c. $\sum_{n=0}^{+\infty} (\frac{1}{2^n} + \frac{1}{3^n})$ d. $\sum_{n=1}^{+\infty} \frac{2n-1}{n^2-3}$ e. $\sum_{n=1}^{+\infty} (\frac{1}{n^2} - \frac{1}{n})$ f. $\sum_{n=2}^{+\infty} \frac{1}{n^2-1}$ 2. Give examples of series $\sum a_n$ and $\sum b_n$ such that: a. $\sum a_n$ and $\sum b_n$ are divergent and $\sum (a_n + b_n)$ is convergent; b. $\sum a_n$ and $\sum b_n$ are divergent and $\sum (a_n + b_n)$ is divergent. ### Convergence Tests for Series of Nonnegative Terms Throughout this section we consider only series $\sum a_n$ where $a_n \geq 0$ for all $n$. #### Theorem 2.3.1 (Bounded-Sum Test). Let $\sum_{n=1}^{+\infty} a_n$ be a series of positive terms. Then $\sum_{n=1}^{+\infty} a_n$ converges if and only if its sequence of partial sums is bounded above. **Proof:** Let $S_n = a_1 + \dots + a_n$. Then for all $n$ we have $0 S_n$. Therefore $\{S_n\}_{n=1}^{+\infty}$ is increasing, and hence is monotonic. Also $S_n > 0$ for all $n$, so $\{S_n\}_{n=1}^{+\infty}$ is bounded below. Thus $\{S_n\}_{n=1}^{+\infty}$ is bounded if and only if it is bounded above. By the Monotone Convergence Theorem, $\{S_n\}_{n=1}^{+\infty}$ is convergent if and only if it is bounded. Putting everything together, we get that $\sum_{n=1}^{+\infty} a_n$ converges if and only if its sequence of partial sums is bounded above. #### Example 2.3.2. Consider the telescoping sum $\sum_{n=1}^{+\infty} (\frac{1}{n} - \frac{1}{n+1})$. Observe that for all $n \in N$, $a_n = \frac{1}{n} - \frac{1}{n+1} > 0$. Moreover, for all $k \in N$, $S_k = \sum_{n=1}^k (\frac{1}{n} - \frac{1}{n+1}) = (1 - \frac{1}{2}) + (\frac{1}{2} - \frac{1}{3}) + \dots + (\frac{1}{k} - \frac{1}{k+1}) = 1 - \frac{1}{k+1}$. Since $0 a_m + \int_1^m f(x) dx$. It follows from the hypothesis that $\lim_{m \to +\infty} (a_m + \int_1^m f(x) dx) = +\infty$, which implies that $\lim_{m \to +\infty} S_m = +\infty$. Hence $\sum_{n=1}^{+\infty} a_n$ diverges. #### Example 2.3.4. Determine whether the series $\sum_{n=1}^{+\infty} \frac{2n}{1+n^2}$ converges or diverges. **Solution:** For $x \geq 1$, let $f(x) = \frac{2x}{1+x^2}$. Clearly, $f$ is positive-valued and continuous on $[1, +\infty)$. Moreover, $f'(x) = \frac{2(1+x^2) - 2x(2x)}{(1+x^2)^2} = \frac{2+2x^2-4x^2}{(1+x^2)^2} = \frac{2-2x^2}{(1+x^2)^2} = \frac{2(1-x^2)}{(1+x^2)^2} \leq 0$ for all $x \geq 1$, so $f$ is decreasing on $[1, +\infty)$. The Integral Test can then be applied. Since $\int_1^{+\infty} \frac{2x}{1+x^2} dx = \lim_{t \to +\infty} \int_1^t \frac{2x}{1+x^2} dx = \lim_{t \to +\infty} [\ln(1+x^2)]_1^t = \lim_{t \to +\infty} (\ln(1+t^2) - \ln(2)) = +\infty$. So the series $\sum_{n=1}^{+\infty} \frac{2n}{1+n^2}$ diverges by the Integral Test. #### Example 2.3.5. Determine whether the series $\sum_{n=2}^{+\infty} \frac{1}{n(\ln n)^2}$ converges or diverges. **Solution:** For $x \geq 2$, let $f(x) = \frac{1}{x(\ln x)^2}$. Clearly, $f$ is positive-valued and continuous on $[2, +\infty)$. Since $x(\ln x)^2$ increases on $[2, +\infty)$, it follows that $f$ decreases on $[2, +\infty)$. Hence the Integral Test can be applied. $\int_2^{+\infty} \frac{1}{x(\ln x)^2} dx = \lim_{t \to +\infty} \int_2^t \frac{1}{x(\ln x)^2} dx$. Let $u = \ln x$, $du = \frac{1}{x} dx$. When $x=2$, $u=\ln 2$. When $x=t$, $u=\ln t$. So $\lim_{t \to +\infty} \int_{\ln 2}^{\ln t} \frac{1}{u^2} du = \lim_{t \to +\infty} [-\frac{1}{u}]_{\ln 2}^{\ln t} = \lim_{t \to +\infty} (-\frac{1}{\ln t} - (-\frac{1}{\ln 2})) = 0 + \frac{1}{\ln 2} = \frac{1}{\ln 2}$. Therefore $\sum_{n=2}^{+\infty} \frac{1}{n(\ln n)^2}$ converges by the Integral Test. #### Example 2.3.6 (Behavior of p-series). Let $p \in R^+$. Find all values of $p$ for which the p-series $\sum_{n=1}^{+\infty} \frac{1}{n^p}$ is convergent. **Solution:** If $p=1$, the series is the harmonic series, which is divergent by Example 2.2.7. Suppose that $p \neq 1$. Let $f(x) = \frac{1}{x^p}$. Then $f$ is positive-valued and continuous on $[1, +\infty)$. Also $f'(x) = -px^{-p-1} 1$. If $p > 1$, then $\lim_{t \to +\infty} t^{-p+1} = 0$, so the integral converges to $0 - \frac{1}{1-p} = \frac{1}{p-1}$. If $0 0$, so $\lim_{t \to +\infty} t^{-p+1} = +\infty$, and the integral diverges to $+\infty$. By the Integral Test, a p-series is convergent if $p>1$ and divergent if $0 \frac{1}{\sqrt{n}}$. The series $\sum_{n=2}^{+\infty} \frac{1}{\sqrt{n}}$ is a p-series with $p = \frac{1}{2} 3^n$ for all $n \in N$, we have $\frac{2^n \sin^2 n}{3^n+1} \leq \frac{2^n}{3^n+1} 1$ and so is convergent. Since for each $n \geq 1$, $\lim_{n \to +\infty} \frac{a_n}{b_n} = \lim_{n \to +\infty} \frac{4n^2+3n-1}{3n^5+n} \cdot n^3 = \lim_{n \to +\infty} \frac{4n^5+3n^4-n^3}{3n^5+n} = \lim_{n \to +\infty} \frac{n^5(4+\frac{3}{n}-\frac{1}{n^2})}{n^5(3+\frac{1}{n^4})} = \frac{4}{3}$. Since $\frac{4}{3} \in R^+$, it follows from the Limit Comparison Test that $\sum_{n=1}^{+\infty} \frac{4n^2+3n-1}{3n^5+n}$ is also convergent. #### Example 2.3.14. Determine whether the series $\sum_{n=2}^{+\infty} \frac{1}{n(\ln n)^2}$ is convergent or divergent. **Solution:** Let $a_n = \frac{1}{n(\ln n)^2}$ and $b_n = \frac{1}{n}$. The series $\sum_{n=2}^{+\infty} \frac{1}{n}$ diverges. $\lim_{n \to +\infty} \frac{a_n}{b_n} = \lim_{n \to +\infty} \frac{1}{n(\ln n)^2} \cdot n = \lim_{n \to +\infty} \frac{1}{(\ln n)^2}$. Using L'Hopital's Rule for $x \to +\infty$, $\lim_{x \to +\infty} \frac{1}{(\ln x)^2} = 0$. By the Limit Comparison Test (part 2), if $\lim_{n \to +\infty} \frac{a_n}{b_n} = 0$ and $\sum b_n$ diverges, the test is inconclusive. **Warning:** The example solution for 2.3.14 incorrectly states it diverges by LCT. The original example 2.3.5 already proved convergence using the Integral Test. #### Exercises 1. Determine whether the given series is convergent or divergent using (i) the Integral Test, (ii) the Comparison Test, and (iii) the Limit Comparison Test. a. $\sum_{n=3}^{+\infty} \frac{\ln n}{n}$ b. $\sum_{n=1}^{+\infty} \sin(\frac{1}{n})$ c. $\sum_{n=1}^{+\infty} \frac{n}{e^{n^2}}$ 2. Consider the series $\sum_{n=3}^{+\infty} \frac{\ln n}{n^2}$. a. Explain why the Comparison Test and Limit Comparison Test fail when applied to the given series compared to the harmonic series. b. Explain why the Comparison Test and Limit Comparison Test fail when the given series is compared to the p-series with $p=2$. c. Determine whether the series is convergent or divergent. 3. Determine whether the series converges or diverges. a. $\sum_{n=2}^{+\infty} \frac{n^3+11}{4n^5-7n^3+3}$ b. $\sum_{n=2}^{+\infty} \frac{1}{\ln n}$ c. $\sum_{n=1}^{+\infty} \sin(\frac{1}{n})$ d. $\sum_{n=1}^{+\infty} e^{-n^2}$ e. $\sum_{n=1}^{+\infty} \frac{1+\cos n}{n^2-\cos n}$ f. $\sum_{n=1}^{+\infty} \coth(\frac{1}{n})$ 4. Use the Integral Test to find all values of $p$ such that $\sum_{n=1}^{+\infty} \frac{1}{n(\ln n)^p}$ converges. 5. Given the series $\sum_{n=3}^{+\infty} \frac{\ln n}{n^p}$. a. Use the Comparison Test to show that the series diverges when $0 2$. b. Use the Limit Comparison Test to show that the series converges when $1 ### Alternating Series Test #### Definition 2.4.1. An **alternating series** is a series of the form $\sum (-1)^n b_n$ or $\sum (-1)^{n-1} b_n$, where $b_n > 0$ for each $n$. This means an alternating series has non-zero terms that alternate in sign. #### Theorem 2.4.2 (Alternating Series Test). Let $\sum (-1)^n b_n$ be an alternating series with $b_n \geq 0$ for all $n$. If (i) the sequence $\{b_n\}$ is ultimately decreasing, and (ii) $\lim_{n \to +\infty} b_n = 0$ then the series is convergent. **Proof:** Consider the series $\sum_{n=1}^{+\infty} (-1)^{n-1} b_n$, where $b_n > 0$ for all $n$, and assume it satisfies conditions (i) and (ii). Let $S_k = b_1 - b_2 + b_3 - \dots + (-1)^{k-1} b_k$. If $k$ is even, say $k=2n$, then $S_{2n} = (b_1 - b_2) + (b_3 - b_4) + \dots + (b_{2n-1} - b_{2n})$. Since $\{b_n\}$ is decreasing, $b_i - b_{i+1} \geq 0$, so each term in parentheses is non-negative. Also $S_{2n} = b_1 - (b_2 - b_3) - \dots - (b_{2n-2} - b_{2n-1}) - b_{2n} \leq b_1$. So $S_{2n+2} = S_{2n} + (b_{2n+1} - b_{2n+2}) \geq S_{2n}$. The sequence $\{S_{2n}\}_{n=1}^{+\infty}$ of partial sums for even $k$ is increasing and bounded above, and hence is monotonic and bounded. By the Bounded-Sum Test, $\{S_{2n}\}_{n=1}^{+\infty}$ is convergent. If $k$ is odd, say $k=2n-1$, then $S_{2n-1} = b_1 - b_2 + \dots + b_{2n-1} = (b_1 - b_2) + (b_3 - b_4) + \dots + (b_{2n-3} - b_{2n-2}) + b_{2n-1} \geq b_1 - b_2$. Also $S_{2n+1} = S_{2n-1} - (b_{2n} - b_{2n+1}) \leq S_{2n-1}$. The sequence $\{S_{2n-1}\}_{n=1}^{+\infty}$ of partial sums for odd $k$ is decreasing and bounded below, and hence is monotonic and bounded. By the Bounded-Sum Test, $\{S_{2n-1}\}_{n=1}^{+\infty}$ is convergent. Finally we show that $\lim_{n \to +\infty} S_{2n} = \lim_{n \to +\infty} S_{2n-1}$. Indeed, $\lim_{n \to +\infty} S_{2n} - \lim_{n \to +\infty} S_{2n-1} = \lim_{n \to +\infty} (S_{2n} - S_{2n-1}) = \lim_{n \to +\infty} a_{2n} = 0$. This implies that $\lim_{n \to +\infty} S_{2n} = \lim_{n \to +\infty} S_{2n-1}$. Therefore the sequence $\{S_k\}$ converges, and hence $\sum_{n=1}^{+\infty} (-1)^{n-1} b_n$ converges. #### Warning: If either $\lim_{n \to +\infty} b_n \neq 0$ or $\{b_n\}$ is not ultimately decreasing, the Alternating Series Test is inconclusive. Other tests must be used to determine convergence or divergence. #### Example 2.4.3. Determine whether the series $\sum_{n=1}^{+\infty} \frac{(-1)^{n-1}}{n}$ converges or diverges. **Solution:** Let $b_n = \frac{1}{n}$, which is positive for each $n$. Then $b_{n+1} = \frac{1}{n+1} 1$, $\frac{n+1}{2n} ### Tests for Absolute Convergence Consider the series $\sum_{n=1}^{+\infty} \frac{\sin n}{n^2} = \frac{\sin 1}{1} + \frac{\sin 2}{4} + \frac{\sin 3}{9} + \frac{\sin 4}{16} + \dots$. Observe that some terms are positive and some are negative. In fact, $\sin 1 > 0$, $\sin 2 > 0$, $\sin 3 > 0$, $\sin 4 1$, it is convergent. By the Comparison Test, $\sum_{n=1}^{+\infty} \frac{\sin n}{n^2}$ is convergent. Thus, the series $\sum_{n=1}^{+\infty} \frac{\sin n}{n^2}$ is absolutely convergent. Furthermore, by the previous theorem, it is convergent. #### Example 2.5.6. Determine whether the series $\sum_{n=1}^{+\infty} \frac{2 \cos n}{3n\sqrt{n}+n}$ is absolutely convergent, conditionally convergent, or divergent. **Warning:** The given series contains positive and negative terms but is not an alternating series. Thus the Alternating Series Test cannot be used. **Solution:** Consider the series of absolute values $\sum_{n=1}^{+\infty} |\frac{2 \cos n}{3n\sqrt{n}+n}|$. Observe that for all $n$, $|\frac{2 \cos n}{3n\sqrt{n}+n}| = \frac{2|\cos n|}{3n^{3/2}+n} \leq \frac{2}{3n^{3/2}+n} 1$, and so is convergent. By the Comparison Test, $\sum_{n=1}^{+\infty} |\frac{2 \cos n}{3n\sqrt{n}+n}|$ is convergent. Therefore $\sum_{n=1}^{+\infty} \frac{2 \cos n}{3n\sqrt{n}+n}$ is absolutely convergent. #### Theorem 2.5.7 (Ratio Test). Let $\sum a_n$ be a series of nonzero terms. 1. If $\lim_{n \to +\infty} |\frac{a_{n+1}}{a_n}| = L 1$ for some $L \in R$ or $\lim_{n \to +\infty} |\frac{a_{n+1}}{a_n}| = +\infty$, then the series is divergent. **Warning:** If $\lim_{n \to +\infty} |\frac{a_{n+1}}{a_n}| = 1$, then the Ratio Test is inconclusive. (See Example 2.5.11.) #### Example 2.5.8. Determine whether the series $\sum_{n=1}^{+\infty} \frac{1 \cdot 3 \cdot 5 \cdot \dots \cdot (2n-1)}{n!}$ is convergent or divergent. **Solution:** Let $a_n = \frac{1 \cdot 3 \cdot 5 \cdot \dots \cdot (2n-1)}{n!}$. Then $a_{n+1} = \frac{1 \cdot 3 \cdot 5 \cdot \dots \cdot (2n-1)(2n+1)}{(n+1)!}$. $|\frac{a_{n+1}}{a_n}| = |\frac{1 \cdot 3 \cdot 5 \cdot \dots \cdot (2n-1)(2n+1)}{(n+1)!} \cdot \frac{n!}{1 \cdot 3 \cdot 5 \cdot \dots \cdot (2n-1)}| = \frac{2n+1}{n+1}$. $\lim_{n \to +\infty} |\frac{a_{n+1}}{a_n}| = \lim_{n \to +\infty} \frac{2n+1}{n+1} = \lim_{n \to +\infty} \frac{n(2+\frac{1}{n})}{n(1+\frac{1}{n})} = 2$. Since $2 > 1$, the given series diverges by the Ratio Test. #### Example 2.5.9. Determine whether the series $\sum_{n=1}^{+\infty} \frac{n!}{n^n}$ is convergent or divergent. **Solution:** Let $a_n = \frac{n!}{n^n}$. Then $a_{n+1} = \frac{(n+1)!}{(n+1)^{n+1}}$. $|\frac{a_{n+1}}{a_n}| = |\frac{(n+1)!}{(n+1)^{n+1}} \cdot \frac{n^n}{n!}| = \frac{(n+1)n!}{(n+1)^{n+1}} \cdot \frac{n^n}{n!} = \frac{n^n}{(n+1)^n} = (\frac{n}{n+1})^n = (\frac{1}{1+\frac{1}{n}})^n$. $\lim_{n \to +\infty} |\frac{a_{n+1}}{a_n}| = \lim_{n \to +\infty} (\frac{1}{1+\frac{1}{n}})^n = \frac{1}{\lim_{n \to +\infty} (1+\frac{1}{n})^n} = \frac{1}{e}$. Since $e^{-1} 0$ we have $\lim_{n \to +\infty} |\frac{1/(n+1)^p}{1/n^p}| = \lim_{n \to +\infty} (\frac{n}{n+1})^p = \lim_{n \to +\infty} (\frac{1}{1+\frac{1}{n}})^p = 1^p = 1$. 1. Consider $\sum a_n$ where $a_n = \frac{1}{n^2}$. It is a p-series with $p=2 > 1$. The series is convergent and absolutely convergent. 2. Consider $\sum a_n$ where $a_n = \frac{(-1)^n}{n}$. It is the alternating harmonic series, which is conditionally convergent. 3. Consider $\sum a_n$ where $a_n = \frac{1}{n}$. It is the harmonic series, which is divergent. #### Theorem 2.5.12 (Root Test). Given a series $\sum a_n$. 1. If $\lim_{n \to +\infty} \sqrt[n]{|a_n|} = L 1$ for some $L \in R$ or $\lim_{n \to +\infty} \sqrt[n]{|a_n|} = +\infty$, then the series is divergent. **Warning:** If $\lim_{n \to +\infty} \sqrt[n]{|a_n|} = 1$, then the Root Test is inconclusive. (See Example 2.5.15.) #### Example 2.5.13. Determine whether the series $\sum_{n=1}^{+\infty} (\frac{2}{\tan^{-1}(n)})^n$ converges or diverges. **Solution:** Let $a_n = (\frac{2}{\tan^{-1}(n)})^n$. $\lim_{n \to +\infty} \sqrt[n]{|a_n|} = \lim_{n \to +\infty} \sqrt[n]{|(\frac{2}{\tan^{-1}(n)})^n|} = \lim_{n \to +\infty} \frac{2}{\tan^{-1}(n)}$. Since $\lim_{n \to +\infty} \tan^{-1}(n) = \frac{\pi}{2}$, the limit is $\frac{2}{\pi/2} = \frac{4}{\pi}$. Since $\frac{4}{\pi} \approx 1.27 > 1$, the given series is divergent by the Root Test. #### Example 2.5.14. Determine whether the series $\sum_{n=1}^{+\infty} \frac{(\ln n)^{2n}}{n^n}$ is convergent or divergent. **Solution:** Let $a_n = \frac{(\ln n)^{2n}}{n^n}$. $\lim_{n \to +\infty} \sqrt[n]{|a_n|} = \lim_{n \to +\infty} \sqrt[n]{|\frac{(\ln n)^{2n}}{n^n}|} = \lim_{n \to +\infty} \frac{(\ln n)^2}{n}$. This limit is of the form $\frac{\infty}{\infty}$, so we can use L'Hopital's Rule for $x \to +\infty$: $\lim_{x \to +\infty} \frac{(\ln x)^2}{x} \overset{L'HR}{=} \lim_{x \to +\infty} \frac{2 \ln x \cdot \frac{1}{x}}{1} = \lim_{x \to +\infty} \frac{2 \ln x}{x} \overset{L'HR}{=} \lim_{x \to +\infty} \frac{2/x}{1} = 0$. Since $\lim_{n \to +\infty} \sqrt[n]{|a_n|} = 0 ### Power Series #### Definition 2.6.1. Let $a \in R$. A **power series about $a$** (or **centered at $a$**) is an expression of the form $\sum_{n=0}^{+\infty} c_n (x-a)^n = c_0 + c_1(x-a) + c_2(x-a)^2 + c_3(x-a)^3 + \dots + c_n(x-a)^n + \dots$, where $x$ is a variable and $c_n$ is a constant for all $n$. #### Remark 2.6.2. 1. **Convention:** $(x-a)^0 := 1$ for all $x \in R$. 2. If there is a largest number $d$ such that $c_d \neq 0$, then the power series is a polynomial with degree $d$. Not all power series are polynomials. #### Example 2.6.3. 1. $\sum_{n=0}^{+\infty} x^n$ is a power series centered at $0$, with $c_n = 1$ for all $n$. 2. $\sum_{n=0}^{+\infty} \frac{(x-1)^n}{n!}$ is a power series centered at $1$, with $c_n = \frac{1}{n!}$ for all $n$. 3. $\sum_{n=1}^{+\infty} \frac{(x+1)^{2n}}{n^2}$ is a power series centered at $-1$. It can be written as $\sum_{k=0}^{+\infty} c_k(x+1)^k$ where $c_k = 0$ if $k$ is odd, and $c_k = \frac{1}{(k/2)^2}$ if $k$ is even. 4. $\sum_{n=0}^{+\infty} (\frac{2-5x}{3})^n$ can be written as $\sum_{n=0}^{+\infty} (\frac{-5}{3})^n (x-\frac{2}{5})^n$, and so is a power series centered at $\frac{2}{5}$, with $c_n = (\frac{-5}{3})^n$. 5. The following are NOT power series: $\sum \frac{1}{x^n}$, $\sum \sin(n)x^n$, $\sum \cos(n\pi)x^n$. #### Definition 2.6.4. The **interval of convergence** of a power series in the variable $x$ is the set of all values of $x$ for which the resulting series converges. #### Remark 2.6.5. Every power series $\sum c_n (x-a)^n$ is convergent when $x=a$. Indeed, $\sum_{n=0}^{+\infty} c_n (a-a)^n = c_0 + \sum_{n=1}^{+\infty} c_n (0)^n = c_0 + 0 = c_0$. #### Theorem 2.6.6. For a given power series $\sum_{n=0}^{+\infty} c_n(x-a)^n$, exactly one of the following is true: 1. The series converges only at $x=a$. 2. The series converges absolutely for all $x \in R$. 3. There is a positive number $R$ such that the series converges absolutely if $|x-a| R$. #### Remark 2.6.7. 1. By Theorem 2.6.6, the interval of convergence of a power series centered at $a$ is one of the following forms: * $\{a\}$; * $(-\infty, +\infty)$; * an interval with endpoints $a-R$ and $a+R$ for some real number $R > 0$. 2. We say that $\sum_{n=0}^{+\infty} c_n(x-a)^n$ has **radius of convergence**: * $0$ if its interval of convergence is $\{a\}$; * $\infty$ if its interval of convergence is $(-\infty, +\infty)$; * $R$ if its interval of convergence is an interval with endpoints $a-R$ and $a+R$ for some real number $R > 0$. 3. In many cases, the radius and interval of convergence can be obtained using the conditions of the Ratio Test. #### Example 2.6.8. Find the radius and interval of convergence of the power series $\sum_{n=0}^{+\infty} x^n$. **Solution:** Observe that the given power series is geometric with common ratio $x$. By Example 2.2.5, the power series converges absolutely if $|x| 1$. Hence, the interval of convergence is $(-1,1)$ and the radius of convergence is $1$. **Another Solution:** For all $n \in N$, let $a_n = x^n$. For $x \neq 0$, $\lim_{n \to +\infty} |\frac{a_{n+1}}{a_n}| = \lim_{n \to +\infty} |\frac{x^{n+1}}{x^n}| = \lim_{n \to +\infty} |x| = |x|$. By the Ratio Test the series is absolutely convergent when $|x| 1$. It is inconclusive when $|x|=1$, i.e., $x=1$ or $x=-1$. #### Example 2.6.9. Find the radius and interval of convergence of the power series $\sum_{n=0}^{+\infty} \frac{x^n}{n!}$. **Solution:** For $n \in N$, let $a_n = \frac{x^n}{n!}$. For $x \neq 0$, $\lim_{n \to +\infty} |\frac{a_{n+1}}{a_n}| = \lim_{n \to +\infty} |\frac{x^{n+1}}{(n+1)!} \cdot \frac{n!}{x^n}| = \lim_{n \to +\infty} |\frac{x}{n+1}| = |x| \lim_{n \to +\infty} \frac{1}{n+1} = |x| \cdot 0 = 0$. Since $0 2$, i.e., $x 3$. Note that the Ratio Test fails if $|x-1|=2$, which corresponds to $x=-1$ and $x=3$. We consider these cases separately: * If $x=3$, the series becomes $\sum_{n=1}^{+\infty} \frac{(3-1)^n}{2^n n^2} = \sum_{n=1}^{+\infty} \frac{2^n}{2^n n^2} = \sum_{n=1}^{+\infty} \frac{1}{n^2}$. This is a p-series with $p=2 > 1$, so it is convergent. * If $x=-1$, the series becomes $\sum_{n=1}^{+\infty} \frac{(-1-1)^n}{2^n n^2} = \sum_{n=1}^{+\infty} \frac{(-2)^n}{2^n n^2} = \sum_{n=1}^{+\infty} \frac{(-1)^n}{n^2}$. This is absolutely convergent (since $\sum \frac{1}{n^2}$ converges), and hence is convergent. Therefore, the interval of convergence is $[-1, 3]$ and the radius of convergence is $2$. #### Example 2.6.11. Find the radius and interval of convergence of $\sum_{n=2}^{+\infty} \frac{(3x-2)^n}{\ln n}$. **Solution:** Let $a_n = \frac{(3x-2)^n}{\ln n}$. Then for $x \neq \frac{2}{3}$, $\lim_{n \to +\infty} |\frac{a_{n+1}}{a_n}| = \lim_{n \to +\infty} |\frac{(3x-2)^{n+1}}{\ln(n+1)} \cdot \frac{\ln n}{(3x-2)^n}| = |3x-2| \lim_{n \to +\infty} \frac{\ln n}{\ln(n+1)}$. Let $f(t) = \ln t$. Then $\lim_{t \to +\infty} \frac{\ln t}{\ln(t+1)} \overset{L'HR}{=} \lim_{t \to +\infty} \frac{1/t}{1/(t+1)} = \lim_{t \to +\infty} \frac{t+1}{t} = 1$. So $\lim_{n \to +\infty} |\frac{a_{n+1}}{a_n}| = |3x-2| \cdot 1 = |3x-2|$. By the Ratio Test, the power series converges absolutely if $|3x-2| 1$, i.e., $x 1$. Now we check the endpoints: * If $x=1$, we get the series $\sum_{n=2}^{+\infty} \frac{(3(1)-2)^n}{\ln n} = \sum_{n=2}^{+\infty} \frac{1^n}{\ln n} = \sum_{n=2}^{+\infty} \frac{1}{\ln n}$. For $n \geq 2$, $\ln n \frac{1}{n}$. Since $\sum_{n=2}^{+\infty} \frac{1}{n}$ is a divergent harmonic series (minus the first term), by the Comparison Test, $\sum_{n=2}^{+\infty} \frac{1}{\ln n}$ is divergent. * If $x=\frac{1}{3}$, the series becomes $\sum_{n=2}^{+\infty} \frac{(3(\frac{1}{3})-2)^n}{\ln n} = \sum_{n=2}^{+\infty} \frac{(-1)^n}{\ln n}$. Let $b_n = \frac{1}{\ln n}$. For $n \geq 2$, $b_n > 0$. Since $\ln n$ is increasing, $\frac{1}{\ln n}$ is decreasing. Also $\lim_{n \to +\infty} \frac{1}{\ln n} = 0$. By the Alternating Series Test, $\sum_{n=2}^{+\infty} \frac{(-1)^n}{\ln n}$ converges. Hence the interval of convergence is $[\frac{1}{3}, 1)$ and the radius of convergence is $\frac{1}{3}$. #### Exercises 1. Find the radius and interval of convergence of the given power series. a. $\sum_{n=0}^{+\infty} \frac{(x-a)^n}{n!}$, for any $a \in R$ b. $\sum_{n=0}^{+\infty} \frac{(x-2)^n}{n!}$ c. $\sum_{n=1}^{+\infty} \frac{(3x+4)^n}{n^2}$ d. $\sum_{n=4}^{+\infty} \frac{(1-2x)^n}{\sqrt{n^2-9}}$ e. $\sum_{n=0}^{+\infty} (-5x)^n$ f. $\sum_{n=0}^{+\infty} n! (x-2)^n$ ### Functions as Power Series The sum of a power series $\sum_{n=0}^{+\infty} c_n(x-a)^n$ is a function $f$ given by $f(x) = \sum_{n=0}^{+\infty} c_n(x-a)^n$ for all $x$ in the interval of convergence of the series. In this section, we'll look at two problems: 1. Given a power series, find a non-series expression for its sum $f(x)$, along with the values of $x$ for which it's valid. 2. Given a non-series expression $g(x)$, find a power series whose sum is $g(x)$, along with the values of $x$ for which it's valid. #### Example 2.7.1. Find a non-series expression for the sum of the power series $\sum_{n=0}^{+\infty} x^n$. **Solution:** This is a geometric series with first term $1$ and common ratio $x$. By Example 2.2.5, this series converges to the sum $\frac{1}{1-x}$ when $|x| ### Taylor and Maclaurin Series We continue our study of finding a power series representation for a given function $g(x)$. Under certain conditions, we can define a power series, called a **Taylor series**, associated with $g(x)$. #### Theorem 2.8.1. Assume that the function $g$ has a power series representation about the number $a \in \text{dom}(g)$. Then the derivatives of $g$ of all orders exist at $a$, and $g(x) = \sum_{n=0}^{+\infty} \frac{g^{(n)}(a)}{n!}(x-a)^n$ for all $x$ in the interval of convergence of the series, where $g^{(0)} := g$. **Proof:** Let $a \in R$ and assume that $g(x) = \sum_{n=0}^{+\infty} c_n(x-a)^n$, where $c_i \in R$ for all $i$. By Theorem 2.7.3, $g'(x) = \sum_{n=1}^{+\infty} n c_n(x-a)^{n-1}$ $g''(x) = \sum_{n=2}^{+\infty} n(n-1) c_n(x-a)^{n-2}$ It can be shown that for any $k \in Z^+$, $g^{(k)}(x) = \sum_{n=k}^{+\infty} n(n-1)\dots(n-k+1) c_n(x-a)^{n-k}$. Thus, if $g^{(0)}(x) := g(x)$, then for any nonnegative integer $k$, $g^{(k)}(a) = k! c_k$. Therefore $g^{(k)}(a)$ exists for all $k$, and $c_k = \frac{g^{(k)}(a)}{k!}$. The result follows. #### Definition 2.8.2. The series in Theorem 2.8.1 is called the **Taylor series** for $g$ about the number $a$. The Taylor series for $g$ about $0$ is called the **Maclaurin series** for $g$. #### Example 2.8.3. Find the Taylor series for $g(x) = \frac{1}{x}$ about $1$, and determine the interval of convergence of the Taylor series. **Solution:** Observe that $g'(x) = -\frac{1}{x^2}$, $g''(x) = \frac{2}{x^3}$, $g'''(x) = -\frac{6}{x^4}$. If $g^{(k)}(x) = (-1)^k \frac{k!}{x^{k+1}}$, then $g^{(k+1)}(x) = (-1)^{k+1} \frac{(k+1)!}{x^{k+2}}$. So $g^{(n)}(1) = (-1)^n n!$ for all nonnegative integers $n$. The Taylor series for $g$ about $1$ is $\sum_{n=0}^{+\infty} \frac{(-1)^n n!}{n!}(x-1)^n = \sum_{n=0}^{+\infty} (-1)^n (x-1)^n$. This series is geometric with common ratio $-(x-1)$. It is convergent when $|-(x-1)| 1$. So the interval of convergence of the series is $(0,2)$. #### Example 2.8.4. Find the Maclaurin series for $e^x$ and its interval of convergence. **Solution:** Let $g(x) = e^x$. Then for all $n$ we have $g^{(n)}(x) = e^x$, so $g^{(n)}(0) = e^0 = 1$. Thus $c_n = \frac{g^{(n)}(0)}{n!} = \frac{1}{n!}$. The Maclaurin series for $e^x$ is $\sum_{n=0}^{+\infty} \frac{x^n}{n!}$. It was shown in Example 2.6.9 that this series has interval of convergence $R = (-\infty, +\infty)$. #### Example 2.8.5. Show that $e^x$ is equal to its Maclaurin series for all $x \in R$. **Proof:** Let $h(x) = \sum_{n=0}^{+\infty} \frac{x^n}{n!}$. By Theorem 2.7.3, $h'(x) = \sum_{n=1}^{+\infty} n \frac{x^{n-1}}{n!} = \sum_{n=1}^{+\infty} \frac{x^{n-1}}{(n-1)!} = \sum_{m=0}^{+\infty} \frac{x^m}{m!} = h(x)$. So $h$ satisfies the differential equation $h'(x) = h(x)$, which implies that $h(x) = C e^x$ for some constant $C$. Letting $x=0$, $h(0) = \sum_{n=0}^{+\infty} \frac{0^n}{n!} = 1$. So $1 = C e^0 = C$. Thus $C=1$. Therefore $h(x) = e^x$, which proves that $e^x = \sum_{n=0}^{+\infty} \frac{x^n}{n!}$, as required. #### Example 2.8.6. Use the result of Example 2.8.5 to find the sum of $\sum_{n=1}^{+\infty} \frac{1}{n!(n+2)}$. **Solution:** Consider the expression $x e^x$. From Example 2.8.5 we have $x e^x = x \sum_{n=0}^{+\infty} \frac{x^n}{n!} = \sum_{n=0}^{+\infty} \frac{x^{n+1}}{n!}$. By Theorem 2.7.3, $\int_0^x t e^t dt = \int_0^x \sum_{n=0}^{+\infty} \frac{t^{n+1}}{n!} dt = \sum_{n=0}^{+\infty} \frac{1}{n!} \int_0^x t^{n+1} dt = \sum_{n=0}^{+\infty} \frac{1}{n!} [\frac{t^{n+2}}{n+2}]_0^x = \sum_{n=0}^{+\infty} \frac{x^{n+2}}{n!(n+2)}$. Also, integrating by parts, $\int_0^x t e^t dt = [t e^t]_0^x - \int_0^x e^t dt = x e^x - [e^t]_0^x = x e^x - e^x + 1 = (x-1)e^x + 1$. Equating the two expressions for $x=1$ (which is valid since the series converges for any $x \in R$), we get $(1-1)e^1+1 = \sum_{n=0}^{+\infty} \frac{1^{n+2}}{n!(n+2)} = \sum_{n=0}^{+\infty} \frac{1}{n!(n+2)}$. So $1 = \frac{1}{0!(0+2)} + \sum_{n=1}^{+\infty} \frac{1}{n!(n+2)} = \frac{1}{2} + \sum_{n=1}^{+\infty} \frac{1}{n!(n+2)}$. Therefore $\sum_{n=1}^{+\infty} \frac{1}{n!(n+2)} = 1 - \frac{1}{2} = \frac{1}{2}$. **Warning:** It is not always true that a function is equal to its Taylor series at all points in the interval of convergence of the series. #### Example 2.8.7. Find the Maclaurin series for $f(x) = \begin{cases} e^{-1/x^2} & \text{if } x \neq 0 \\ 0 & \text{if } x = 0 \end{cases}$ and show that $f(x)$ is equal to its Maclaurin series if and only if $x=0$. **Solution:** Observe that $\lim_{x \to 0} e^{-1/x^2} = 0$, so $f$ is continuous at $x=0$. Also $f'(0) = \lim_{x \to 0} \frac{f(x)-f(0)}{x-0} = \lim_{x \to 0} \frac{e^{-1/x^2}-0}{x-0} = \lim_{x \to 0} \frac{e^{-1/x^2}}{x}$. This limit is $0$. (details left as exercise). It can also be shown that for any positive integer $n$, the $n^{th}$ derivative of $e^{-1/x^2}$ with respect to $x$ is of the form $p_n(\frac{1}{x})e^{-1/x^2}$, where $p_n$ is a polynomial. Thus $f^{(n+1)}(0) = \lim_{x \to 0} \frac{f^{(n)}(x)-f^{(n)}(0)}{x-0} = 0$. (details left as exercise). So the Maclaurin series of $f(x)$ is $\sum_{n=0}^{+\infty} \frac{0}{n!} x^n = \sum_{n=0}^{+\infty} 0 = 0$, which converges to $0$ for all $x \in R$. However $f(x) \neq 0$ whenever $x \neq 0$. Therefore $f(x)$ is equal to its Maclaurin series if and only if $x=0$. #### Example 2.8.8. Find the Maclaurin series for $\sin x$. **Solution:** Let $g(x) = \sin x$. The derivatives of $\sin x$ repeat after a cycle of four: $g^{(n)}(x) = \begin{cases} \sin x & \text{if } n=4k \\ \cos x & \text{if } n=4k+1 \\ -\sin x & \text{if } n=4k+2 \\ -\cos x & \text{if } n=4k+3 \end{cases}$. At $x=0$, the non-zero derivatives are those with odd orders: $g^{(n)}(0) = \begin{cases} 1 & \text{if } n=4k+1 \\ -1 & \text{if } n=4k+3 \\ 0 & \text{otherwise} \end{cases}$. Hence, the Maclaurin series for $\sin x$ is $\sum_{n=0}^{+\infty} \frac{(-1)^n}{(2n+1)!} x^{2n+1} = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \dots$. In previous examples, we've obtained Maclaurin series for $e^x$, $\ln(1+x)$, and $\sin x$. The table below shows the Maclaurin series of some common functions and their radius of convergence. | $g(x)$ | Maclaurin series | Radius of convergence | | :------------- | :------------------------------------------------------- | :-------------------- | | $e^x$ | $\sum_{n=0}^{+\infty} \frac{x^n}{n!}$ | $\infty$ | | $\ln(1+x)$ | $\sum_{n=0}^{+\infty} \frac{(-1)^n}{n+1} x^{n+1}$ | $1$ | | $\sin x$ | $\sum_{n=0}^{+\infty} \frac{(-1)^n}{(2n+1)!} x^{2n+1}$ | $\infty$ | | $\cos x$ | $\sum_{n=0}^{+\infty} \frac{(-1)^n}{(2n)!} x^{2n}$ | $\infty$ | | $\sinh x$ | $\sum_{n=0}^{+\infty} \frac{x^{2n+1}}{(2n+1)!}$ | $\infty$ | | $\cosh x$ | $\sum_{n=0}^{+\infty} \frac{x^{2n}}{(2n)!}$ | $\infty$ | | $(1+x)^m, m \in R$ | $\sum_{n=0}^{+\infty} \frac{m(m-1)\dots(m-n+1)}{n!} x^n$ | $1$ if $m \notin N$, $\infty$ if $m \in N$ | #### Exercises 1. Find the Maclaurin series for $\cos x$ using two methods: a. using Theorem 2.8.1; and b. by differentiating the Maclaurin series of $\sin x$. (It will be shown in the next section that $\sin(x)$ is equal to its Maclaurin series for all $x \in R$.) i. By differentiating both sides of the given equation, show that $\cos 2x = \sum_{n=0}^{+\infty} \frac{(-1)^n 2^{2n}}{(2n)!} x^{2n}$. ii. Use the identity $\sin^2 x = \frac{1}{2}(1-\cos 2x)$ to find a power series representation for $\sin^2 x$. 2. Find the Taylor series of $e^x$ about any $a \in R$ using two methods: a. using Theorem 2.8.1; and b. by writing $e^x = e^a e^{x-a}$ and using Example 2.8.5. 3. Use Example 2.8.5 to find the sum of $\sum_{n=0}^{+\infty} \frac{1}{n!}$ and of $\sum_{n=0}^{+\infty} \frac{(-1)^n}{n!}$. 4. Express $t^3 e^{-2t^2} dt$ as a power series and find its interval of convergence. 5. Derive the Maclaurin series for $\sinh x$ and $\cosh x$ and compute the radius of convergence of each series. 6. a. Show that $(1+x)^m$, where $m$ is not a nonnegative integer, is equal to its Maclaurin series for all $x \in (-1,1)$. (Hint: Let $h(x) = \sum_{n=0}^{+\infty} \frac{m(m-1)\dots(m-n+1)}{n!} x^n$. Show that $(1+x)h'(x) = mh(x)$ and solve for $h$.) b. Find the Maclaurin series for $\frac{1}{\sqrt{1-x^2}}$. c. Find the Maclaurin series for $\sin^{-1}(x)$. (Hint: Recall that $D_x \sin^{-1}(x) = \frac{1}{\sqrt{1-x^2}}$). d. Give a series representation for $\pi$ using the Maclaurin series for $\sin^{-1}(x)$. ### Approximations Using Taylor Polynomials In this section, we consider whether a given function is equal to its Taylor series. For functions that are equal to their Taylor series, we estimate function values using partial sums of the Taylor series and determine the accuracy of these approximations. #### Definition 2.9.1. Let $f$ be a function whose first $k$ derivatives exist at $a$. The $k^{th}$ degree **Taylor polynomial of $f$ about $a$** is the polynomial $P_k$ defined by $P_k(x) = \sum_{n=0}^k \frac{f^{(n)}(a)}{n!} (x-a)^n$. The **remainder** is the function $R_k$ given by $R_k(x) = f(x) - P_k(x)$. #### Remark 2.9.2. A Taylor polynomial of $f$ is a partial sum of the Taylor series of $f$. Hence $f(x) = \sum_{n=0}^{+\infty} \frac{f^{(n)}(a)}{n!} (x-a)^n = P_k(x) + R_k(x)$. #### Theorem 2.9.3. Let $f$ be a function whose derivatives of all orders exist at $a$, and let $P_k(x)$ denote the Taylor polynomial of $f$ about $a$. Then $f(x)$ is equal to its Taylor series about $a$ if and only if $\lim_{k \to +\infty} R_k(x) = 0$. **Proof:** By (2.1) and the definition of the remainder function, $f(x) = \sum_{n=0}^{+\infty} \frac{f^{(n)}(a)}{n!} (x-a)^n = \lim_{k \to +\infty} P_k(x)$. So $f(x) = \lim_{k \to +\infty} (f(x) - R_k(x)) = f(x) - \lim_{k \to +\infty} R_k(x)$. It follows immediately that $f(x) = \sum_{n=0}^{+\infty} \frac{f^{(n)}(a)}{n!} (x-a)^n$ exactly when $\lim_{k \to +\infty} R_k(x) = 0$. It's often hard to compute $\lim_{k \to +\infty} R_k(x)$ using its definition. Theorem 2.9.4 and its immediate consequence Corollary 2.9.7 provide indirect methods. #### Theorem 2.9.4 (Lagrange form of $R_k$). Let $f$ be a function whose first $k+1$ derivatives exist in the open interval $I$ between $x$ and $a$ with $f^{(k)}$ continuous in the closed interval between $x$ and $a$. Then for some $z \in I$, with $z$ dependent on $k$, $R_k(x) = \frac{f^{(k+1)}(z)}{(k+1)!} (x-a)^{k+1}$. #### Remark 2.9.5. Theorem 2.9.4 is a generalization of the Mean-Value Theorem for functions learned in Math 21. Recall that if $f$ is a function continuous on $[a,b]$ and differentiable on $(a,b)$, then there exists $z \in (a,b)$ satisfying $f'(z) = \frac{f(b)-f(a)}{b-a}$. This can be written as $f(b)-f(a) = \frac{f'(z)}{1!}(b-a)$. Observe that the left side is equal to $f(b)-P_0(b)$, implying that the right side is $R_0(b)$, which is a special case of (2.2). The following is a frequently useful fact: Since $\sum_{k=0}^{+\infty} \frac{(x-a)^k}{k!}$ is absolutely convergent for all $a, x \in R$, $\lim_{k \to +\infty} \frac{(x-a)^k}{k!} = 0$ for any $a, x \in R$. #### Example 2.9.6. Show that $e^x$ is equal to its Maclaurin series for all $x \in R$. **Proof:** By Theorem 2.9.4, $R_k(x) = \frac{e^z}{(k+1)!} x^{k+1}$ for some $z$ between $0$ and $x$, where $z$ depends on $k$. Clearly $e^0 = 1 = 1+\sum_{n=1}^{+\infty} \frac{1}{n!}$, so we only need to consider $x \neq 0$. We divide our analysis into two cases. **Case 1:** Suppose that $x > 0$. Then $0 0$. **Case 2:** Suppose that $x 0$ and $b_{n+1} ### Chapter Exercises #### I. Suppose that a series $\sum_{n=1}^{+\infty} a_n$ has the property that for all $k \in \{1, 2, 3, \dots\}$, $S_k = \frac{k^2+k}{k^2+\cos k}$. 1. Does the series $\sum_{n=1}^{+\infty} a_n$ converge? If it converges, find its sum. 2. Does the sequence $\{a_n\}_{n=1}^{+\infty}$ converge? If it converges, find its limit. 3. Does the series $\sum_{k=1}^{+\infty} S_k$ converge? If it converges, find its sum. #### II. Let $a_n = \frac{1}{n\sqrt{\ln n}}$ for $n \in N$ with $n \geq 2$. 1. Determine whether $\{a_n\}_{n=2}^{+\infty}$ converges or diverges. 2. Use the Integral Test to determine whether the series $\sum_{n=2}^{+\infty} a_n$ converges or diverges. 3. Show that the series $\sum_{n=2}^{+\infty} (-1)^n a_n$ is conditionally convergent. #### III. Given the series $\sum_{n=1}^{+\infty} \frac{e^n}{n^2}$. Use the Integral Test to determine whether the series converges or diverges. #### IV. Given the series $\sum_{n=1}^{+\infty} \frac{1}{\sqrt{n}}$. 1. Explain why the Comparison Test fails for the given series when it is compared with the series $\sum_{n=1}^{+\infty} \frac{1}{n}$. 2. Use the Comparison Test to determine if the given series converges or diverges. #### V. Determine if the given series converges or diverges. 1. $\sum_{n=1}^{+\infty} \sin(\tan^{-1} n)$ 2. $\sum_{n=1}^{+\infty} \frac{8^n}{3^{2n+1}+n^3}$ 3. $\sum_{n=1}^{+\infty} \frac{(-1)^n 2^n}{2n^2}$ 4. $\sum_{n=1}^{+\infty} \frac{n \cos n}{n^3+n}$ 5. $\sum_{n=1}^{+\infty} (-1)^{n+1} \sin(\frac{1}{n})$ 6. $\sum_{n=2}^{+\infty} \frac{(-1)^n}{n(\ln n)^n}$ 7. $\sum_{n=1}^{+\infty} \frac{1^2 \cdot 2^2 \cdot 3^2 \cdot \dots \cdot n^2}{(2n)!}$ 8. $\sum_{n=2}^{+\infty} (-1)^n (\frac{n^2+1}{n^2-1})^n$ 9. $\sum_{n=1}^{+\infty} (\frac{n}{n+1})^{7/5n}$ #### VI. Show that the following series are conditionally convergent. 1. $\sum_{n=1}^{+\infty} \frac{(-1)^n}{\sqrt{n+5}}$ 2. $\sum_{n=1}^{+\infty} \frac{(-1)^n 2^n}{2^{n+1}+1}$ 3. $\sum_{n=1}^{+\infty} \frac{(-1)^n 2n^2}{2n^3+1}$ #### VII. Find the interval of convergence of the following power series. 1. $\sum_{n=0}^{+\infty} \frac{x^{4n}}{(2n)!}$ 2. $\sum_{n=1}^{+\infty} \frac{(x+1)^{2n}}{n}$ 3. $\sum_{n=0}^{+\infty} \frac{(2x+e)^n}{e^n+1}$ #### VIII. Consider the power series $\frac{1}{1-x} = \sum_{n=0}^{+\infty} x^n$, where $|x|