I would like to prove convergence of the following series: $\sum_{n=1}^\infty {(-1)^n\cdot \arctan\left(\frac{n}{1+n^2}\right)}$ I would like to prove the following series: 
$$\sum_{n=1}^\infty {(-1)^n\cdot \arctan\left(\frac{n}{1+n^2}\right)} $$ 
is convergent (absolutely?) or divergent. I think $\arctan\left(\frac{n}{1+n^2}\right)$ is divergent, but I don't know how it interacts with $(-1)^n$ and how to prove it.
Any ideas would be greatly appreciated.
 A: Your initial series is convergent. To prove it, you may just use the alternating series test:


*

*the function $x \mapsto \arctan \left(\dfrac{x}{1+x^2}\right)$ is decreasing over $[1,\infty)$, since its derivative is negative over $[1,\infty)$: $\left(\arctan \left(\dfrac{x}{1+x^2}\right)\right)'= \dfrac{1-x^2}{1+3 x^2+x^4}\leq 0$,


and


*

*as  $x \to \infty$, you have $\arctan \left(\dfrac{x}{1+x^2}\right)\sim \dfrac1x \longrightarrow 0.$

A: Leibniz's criterion for alternating series works here: $\dfrac n{1+n^2}$ decreases to $0$, and $\arctan x$ is continuous increasing, hence $\arctan \dfrac n{1+n^2}$ decreases (to $0$).
It is not absolutely convergent, because 
$$\frac n{1+n^2}=\frac1n\cdot\frac1{1+\cfrac1{n^2}}=\frac1n+o\Bigl(\frac1n\Bigr)$$
Now $\arctan u=u+o(u)$, so
$$\arctan\frac n{1+n^2}=\frac1n+o\Bigl(\frac1n\Bigr)\sim_\infty \frac1n,$$
which  diverges.
A: To show that the terms are decreasing:
$\arctan\left(\frac{n}{1+n^2}\right)- \arctan\left(\frac{n+1}{1+(n+1)^2}\right)
=\arctan\left(\dfrac{\frac{n}{1+n^2}-\frac{n+1}{1+(n+1)^2}}{1+\frac{n}{1+n^2}\frac{n+1}{1+(n+1)^2}}\right)
$
and
$\dfrac{n}{1+n^2}-\dfrac{n+1}{1+(n+1)^2}
=\dfrac{n(1+(n+1)^2)-(n+1)(1+n^2)}{(1+n^2)(1+(n+1)^2)}
$
and
$\begin{array}\\
n(1+(n+1)^2)-(n+1)(1+n^2)
&=n(n^2+2n+2)-(n^3+n^2+n+1)\\
&=n^3+2n^2+2n-(n^3+n^2+n+1)\\
&=n^2+n-1\\
&> 0\\
\end{array}
$
