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

\(\sum_{n=1}^\infty\frac{(-1)^n}{n^2}=\sum_{n=1}^\infty(-1)^nb_n\)

\(b_n=\frac{1}{n^2}\)

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