Consider the series sum_{n=1}^inftyfrac{(-1)^n}{n^2} a) Show the series converges or diverges using the alternating series test. b) Approximate the sum using the 4-th partial sum(S_4) of the series. c) Calculate the maximum error between partial sum(S_4) and the sum of the series using the remainder term portion of the alternating series test.

Question
Series
asked 2021-03-07
Consider the series \(\sum_{n=1}^\infty\frac{(-1)^n}{n^2}\)
a) Show the series converges or diverges using the alternating series test.
b) Approximate the sum using the 4-th partial sum(\(S_4\)) of the series.
c) Calculate the maximum error between partial sum(\(S_4\)) and the sum of the series using the remainder term portion of the alternating series test.

Answers (1)

2021-03-08
Given series is
\(\sum_{n=1}^\infty\frac{(-1)^n}{n^2}=\sum_{n=1}^\infty(-1)^nb_n\)
\(b_n=\frac{1}{n^2}\)
Since,
\(n^2<(n+1)^2\forall n\in N\)</span>
\(\frac{1}{n^2}>\frac{1}{(n+1)^2}\forall n\in N\)
\(b_n>b_{n+1}\forall n\in N\)
Therefore, sequence \(\left\{b_n\right\}\) is decreasing sequence
Also,
\(\lim_{n\to\infty}b_n=\lim_{n\to\infty}\frac{1}{n^2}=0\)
Therefore, Alternating series test, given series converges.
b) 4th partial sum of the series is evaluated as follows
\(\sum_{n=1}^\infty a_n=\sum_{n=1}^\infty\frac{(-1)^n}{n^2}\)
\(a_n=\frac{(-1)^n}{n^2}\)
\(S_4=a_1+a_2+a_3+a_4\)
\(S_4=\frac{(-1)^1}{1^2}+\frac{(-1)^2}{2^2}+\frac{(-1)^3}{3^2}+\frac{(-1)^4}{4^2}\)
\(=-1+\frac{1}{4}-\frac{1}{9}+\frac{1}{16}=-0.7986\)
c) By Alternating series error estimation theorem,, Maximum error between \(S_4\) and sum S of the series is
\(|S-S_4|\leq|a_5|\)
\(|S-S_4|\leq|\frac{(-1)^5}{5^2}|=0.04\)
Maximum error between \(S_4\) and sum S of the series is 0.04
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