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I have to show that the series $\sum^\infty_{n=1}(-1)^n\frac{n}{n^2+1}$ is conditionally convergent.

I am first going to show the series is convergent by the alternating series which states that a series $\sum^\infty_{n=1}(-1)^ka_k$ converges if $a_k\ge a_{k+1}>0$ and $lim_{k\to\infty}a_k=0.$

Let's choose $a_n=\frac{n}{n^2+1}$. Now $$lim_{n\to\infty}a_n=lim_{n\to\infty}\frac{n}{n^2+1}=lim_{n\to\infty}\frac{1}{n+\frac{1}{n}}=0.$$

However, I don't know how to prove $a_n\ge a_{n+1}>0$. This is what I tried: For every $n\ge1$, $$n^2+1\le (n+1)^2+1 $$ but I don't know how to get to $$\iff \frac{n}{n^2+1}\ge \frac{n+1}{(n+1)^2+1}.$$

I also don't know how to prove the conditional convergence since the absolute value of the series seems to converge instead of diverging.

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Just check that $n[(n+1)^{2}+1]\geq (n^{2}+1)(n+1)$ by expanding the square and multiplying out.

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If $f(x)=\frac x{x^2+1}$, then $f'(x)=\frac{1-x^2}{(x^2+1)^2}$, which is $0$ in $1$ and negative in $(1,\infty)$. So, your sequence is decreasing.

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Senond part,: $\lim_{n\rightarrow\infty} [n/(n^2+1)]/[1/n]=1. $ So divergence of $\sum 1/n$ implies the divergence of $\sum n/n^2+1.$ Hence you series is conditionally convergent.

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Option:

$n\ge 1$;

0)Show that $b_n= n+1/n$ is strictly increasing.

$b_{n+1}-b_n= 1+1/(n+1)-1/n=$

$1- 1/n(n+1)>0$;

Then $a_n:= 1/b_n>0$ is strictly decreasing .

1) $\lim_{n \rightarrow \infty} a_n=0;$

2) Leibniz alternating series test.

https://en.m.wikipedia.org/wiki/Alternating_series_test

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