Is $\sum_{n=1}^\infty \frac{(-1)^n}{n}i^{n(n+1)}$ convergent? How to analyse the convergence of the series
$$\sum_{n=1}^\infty \frac{(-1)^n}{n}i^{n(n+1)}=1-\frac{1}{2}-\frac{1}{3}+\frac{1}{4}+\frac{1}{5}-\frac{1}{6}-\frac{1}{7}+\frac{1}{8}+\frac{1}{9}-\frac{1}{10}-\frac{1}{11}+\dots?$$
I've thought about using results about alternating series but this one is not exactly of one those.  I've also tried to show that the sequence of partial sums is a Cauchy sequence and then to find a convergent subsequence of them, but I wasn't successful either. 
Remark: I even know this series is really convergent...

Note that for any positive integer $N$,
$$S_{2N+1}=\sum_{n=1}^{2N+1} \frac{(-1)^n}{n}i^{n(n+1)}=1+\sum_{n=1}^{N}(-1)^n\left(\frac{1}{2n}+\frac{1}{2n+1}\right).$$
Moreover $S_{2N}=S_{2N+1}-\frac{(-1)^N}{2N+1}$ which implies that if $S_{2N+1}\to L$ then $S_{2N}\to L$ too. Hence the series is convergent by Leibniz. 
Once convergence is established, it follows that
\begin{align*}\sum_{n=1}^\infty \frac{(-1)^n}{n}i^{n(n+1)}&=\lim_{N\to +\infty} S_{2N+1}=
\lim_{N\to +\infty}\left(\frac{1}{2}\sum_{n=1}^{N}\frac{(-1)^n}{n}
+\sum_{n=0}^{N}\frac{(-1)^n}{2n+1}\right)\\
&=\frac{1}{2}\sum_{n=1}^{\infty}\frac{(-1)^n}{n}
+\sum_{n=0}^{\infty}\frac{(-1)^n}{2n+1}=-\frac{\ln(2)}{2}+\frac{\pi}{4}.
\end{align*}
