Does the series $\sum_{n=1}^{\infty} (-1)^n \arctan(n^2)$ converge? I know I can't prove this one with Leibniz, because $\displaystyle \lim_{n\to\infty} \arctan(n^2) = \frac \pi 2 \ne 0 $.
With which kind of test should I prove it? 
EDIT: Okay thanks, Maple calculated and gave the result of -.1729004752 but I guess that's not right then?
 A: By writing the LHS as the argument of an infinite product, then using the Weierstrass product for the sine and cosine function, we have:
$$\sum_{n=1}^{+\infty}(-1)^n \arctan\frac{1}{n^2}=-\arctan\left(\frac{\sinh\frac{\pi}{\sqrt{2}}-\sin\frac{\pi}{\sqrt{2}}}{\sinh\frac{\pi}{\sqrt{2}}+\sin\frac{\pi}{\sqrt{2}}}\right)=-0.6124976882\ldots\tag{1}$$
hence the value returned by Maple is just the Césaro sum of the series, namely:
$$\begin{eqnarray*}{\sum_{n=1}^{+\infty}}^*(-1)^n \arctan n^2&=&\arctan\left(\frac{\sinh\frac{\pi}{\sqrt{2}}-\sin\frac{\pi}{\sqrt{2}}}{\sinh\frac{\pi}{\sqrt{2}}+\sin\frac{\pi}{\sqrt{2}}}\right)-\frac{\pi}{4}\\&=&-\arctan\frac{\sin(\pi/\sqrt{2})}{\sinh(\pi/\sqrt{2})}\\&=&-0.17290047519068984988\ldots.\tag{2}\end{eqnarray*}$$
It is a common issue of many Computer Algebra Systems to incorrectly deal with non-converging series, returning the Césaro sum instead of a warning. I remember old versions of Mathematica giving
$$\sum_{n=0}^{+\infty}\cos(\pi n) = \frac{1}{2},$$
for instance, so I am not surprised of this strange behaviour.
A: Since 
$$\lim_{n\to\infty}\arctan(n^2)=\frac\pi2$$
then the general term of the series doesn't tend to $0$ so the series is divergent.
A: You've just demonstrated that it doesn't converge, because it fails the limit test.
