# 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 su

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.

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Nathanael Webber

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,
$$n^2<(n+1)^2\forall n\in N$$
$$\frac{1}{n^2}>\frac{1}{(n+1)^2}\forall n\in N$$
$$b_n>b_{n+1}\forall n\in N$$
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