# Show that the series $\sum^\infty_{n=1}(-1)^n\frac{n}{n^2+1}$ is conditionally convergent

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.

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

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.

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.

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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