<s>[INST] 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.
 [/INST] 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}
$
</s>