9.4 Alternating Series and Error Bound
Alternating Series Test
Let $a_n>0$. The alternating series
$
\sum_{n=1}^{\infty}(-1)^n a_n \text { and } \sum_{n=1}^{\infty}(-1)^{n+1} a_n
$
converge if the following two conditions are met.
1. $\lim _{n \rightarrow \infty} a_n=0$
2. $a_{n+1} \leq a_n$, for all $n$ greater than some integer $N$.
Alternating Series Estimation Theorem (Error Bound)
If $S_n$ is a partial sum and $S=\sum_{n=1}^{\infty}(-1)^n a_n$ is the sum of a convergent alternating series that satisfies the condition $a_{n+1} \leq a_n$, then the remainder $R_n=S-S_n$ is smaller than $a_{n+1}$, which is the absolute value of the first neglected term.
$
\left|R_n\right|=\left|S-S_n\right| \leq a_{n+1}
$
Definition of Absolute and Conditional Convergence
1. $\sum a_n$ is absolutely convergent if $\sum\left|a_n\right|$ converges.
2. $\sum a_n$ is conditionally convergent if $\sum a_n$ converge but $\sum\left|a_n\right|$ diverges.
Example 1
- Determine whether the series is convergent or divergent.
(a) $\sum_{n=1}^{\infty} \frac{(-1)^n}{\sqrt{n}}$
(b) $\sum_{n=1}^{\infty}(-1)^n \frac{n}{2 n-1}$
Answer/Explanation
Solution
(a) 1. $\lim _{n \rightarrow \infty} a_n=\lim _{n \rightarrow \infty} \frac{1}{\sqrt{n}}=0$
2. $a_{n+1}=\frac{1}{\sqrt{n+1}}<\frac{1}{\sqrt{n}}=a_n$
So the series is convergent by the Alternating Series Test.
(b) 1 . $\lim _{n \rightarrow \infty} a_n=\lim _{n \rightarrow \infty} \frac{n}{2 n-1}=\frac{1}{2} \neq 0$
So the series is divergent by the $n$th Term Test for Divergence.
Example 2
- Determine whether the series is absolutely convergent, conditionally convergent, or divergent.
(a) $\sum_{n=1}^{\infty} \frac{(-1)^n \sqrt[n]{e}}{n^2}$
(b) $\sum_{n=1}^{\infty}(-1)^{n+1} n^{-2 / 3}$
▶️Answer/Explanation
Solution
(a) Since $0 \leq \frac{\sqrt[n]{e}}{n^2} \leq \frac{e}{n^2}=e\left(\frac{1}{n^2}\right)$ and $\sum_{n=1}^{\infty} \frac{1}{n^2}$ is a convergent $p$ – series $(p=2>1)$, $\sum_{n=1}^{\infty} \frac{\sqrt[n]{e}}{n^2}$ converges, and so $\sum_{n=1}^{\infty} \frac{(-1)^n \sqrt[n]{e}}{n^2}$ is absolutely convergent.
(b) $1 . \lim _{n \rightarrow \infty} a_n=\lim _{n \rightarrow \infty} \frac{1}{n^{2 / 3}}=0$
2. $a_{n+1}=\frac{1}{(n+1)^{2 / 3}}<\frac{1}{(n)^{2 / 3}}=a_n$
So the series $\sum_{n=1}^{\infty}(-1)^{n+1} n^{-2 / 3}$ is convergent by the Alternating Series Test.
Now consider the series of absolute values.
$\sum_{n=1}^{\infty}\left|(-1)^{n+1} n^{-2 / 3}\right|=\sum_{n=1}^{\infty} n^{-2 / 3}$ is a divergent $p$ – series $\left(p=\frac{2}{3}<1\right)$.
Thus, $\sum_{n=1}^{\infty}(-1)^{n+1} n^{-2 / 3}$ is conditionally convergent.
Example 3
- Let $f(x)=1-\frac{x^2}{2 !}+\frac{x^4}{4 !}-\frac{x^6}{6 !}+\cdots+-\frac{(-1)^n x^{2 n}}{(2 n) !}+\cdots$
Use the alternating series error bound to show that $1-\frac{1}{2 !}+\frac{1}{4 !}$ approximates $f(1)$ with an error less than $\frac{1}{500}$.
▶️Answer/Explanation
Solution
$
f(1)=1-\frac{1}{2 !}+\frac{1}{4 !}-\frac{1}{6 !}+\cdots+\frac{(-1)^n}{(2 n) !}+\cdots
$
Since series is alternating, with terms convergent to 0 and decreasing in absolute value, the error is less than the first neglected term.
So, $\left|f(1)-\left(1-\frac{1}{2 !}+\frac{1}{4 !}\right)\right| \leq \frac{1}{6 !}=\frac{1}{720}<\frac{1}{500}$.
Example 4
Which of the following series converge?
I. $\sum_{n=1}^{\infty} \frac{(-1)^n \sqrt{n}}{n}$
II. $\sum_{n=1}^{\infty} \frac{(-1)^n}{\ln n}$
III. $\sum_{n=1}^{\infty} \cos (n \pi)$
(A) I only
(B) II only
(C) III only
(D) I and II only
▶️Answer/Explanation
Ans:D
Example 4
Which of the following series converge?
I. $\sum_{n=1}^{\infty}(-1)^n \cos \left(\frac{\pi}{n}\right)$
II. $\sum_{n=1}^{\infty} \sin \left(\frac{2 n-1}{2}\right) \pi$
III. $\sum_{n=1}^{\infty}(-1)^{n+1} \frac{2 n}{n^2+1}$
(A) I only
(B) II only
(C) III only
(D) I and II only
▶️Answer/Explanation
Ans:C
Example 5
For what integer $k, k>1$, will both $\sum_{n=1}^{\infty} \frac{(-1)^{k n}}{\sqrt{n}}$ and $\sum_{n=1}^{\infty} \frac{n^2 \sqrt{n}}{n^k+1}$ converge?
(A) 3
(B) 4
(C) 5
(D) 6
▶️Answer/Explanation
Ans:C
Example6.
- Let $s=\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n^3}$ and $s_n$ be the sum of the first $n$ terms of the series. If $\left|s-s_n\right|<\frac{1}{500}$ what is the smallest value of $n$ ?
(A) 6
(B) 7
(C) 8
(D) 9
Ans:
Example7.
Which of the following series converge?
I. $\sum_{n=2}^{\infty}(-1)^n \sqrt[n]{3}$
II. $\sum_{n=1}^{\infty} \frac{3^{n+1}}{\pi^n}$
III. $\sum_{n=1}^{\infty}\left(\tan ^{-1}(n+1)-\tan ^{-1}(n)\right)$
(A) I only
(B) II only
(C) III only
(D) II and III only
Ans: