Does $\sum_{n=1}^{\infty} \frac{(-1)^n}{n}i^{n(n+1)}$ converge? I was trying to prove or disprove the convergence of the sum $\sum_{n=1}^{\infty} \frac{(-1)^n}{n}i^{n(n+1)}$, but I couldn't find a criterion or method to do it. My intuition leads me to belive that it converges because it is very similar to an the alternating sum, however, $\frac{i^{n(n+1)}}{n}$ is not positive decreasing. Maybe there is a similar criterion to alternating sums that takes this example into account?
 A: The sum for even $n$ is alternating.
The sum for odd $n$ is alternating.
So the sum to $2n$ terms is $A_n+B_n$, and both $A_n$ and $B_n$ converge.
A: This can be rewritten:
$$\sum_{n=1}^{\infty}\frac{(-1)^{n(n+3)/2}}{n}$$
which can be rewritten:
$$\frac{1}{1}-\frac{1}{2}-\frac{1}{3}+\frac{1}{4}+\frac{1}{5}-\frac{1}{6}-\frac{1}{7}+\frac{1}{8}+\cdots +\frac{1}{4k-3}-\frac{1}{4k-2}-\frac{1}{4k-1}+\frac{1}{4k}+\cdots$$
Now, show $\frac{1}{4k-3}-\frac{1}{4k-2}-\frac{1}{4k-1}+\frac{1}{4k}\sim \frac{1}{16k^3}.$ So the series converges.
I think you can prove a general rule: Given $c_0,\cdots,c_{m-1}$, the series:
$$\sum_{n=1}^{\infty} \frac{c_{n\bmod m}}{n}$$ converges if any only if $$\sum_{i=0}^{m-1} c_i=0.$$

If you define $f(z)=\sum_{n=1}^{\infty} \frac{z^n}{n}$ then your sum can be seen to be:
$$\frac{1-i}{2}f(i)+\frac{1+i}{2}f(-i)$$
which is the real part of $(1-i)f(i).$
But we have that $f(z)=-\log(1-z)$ when $|z|<1$ and when $|z|=1$ and $z\neq 1.$ 
So $$f(i)=-\log(1-i) = -\left(\log \sqrt 2-\frac{\pi}{4}i\right)$$
So we get the real part of  $(1-i)f(i)$ is $$\frac{\pi}{4}-\frac{1}{2}\log 2$$
A: This series does converge to $\frac{1}{4} (\pi -\log (4))$. The partial sums oscillate, with smaller and smaller oscillations as $n\to\infty$.
Mathematica gives the partial sums as $\displaystyle\sum_{n=1}^{m} \frac{(-1)^n}{n}i^{n(n+1)}$ as $$\frac{1}{4} \left(-(2-2 i) (-i)^m \Phi (-i,1,m+1)-(2+2 i) i^m \Phi (i,1,m+1)+\pi -\ln (4)\right)$$
Where $\Phi (z,s,a)$ is the "Lerch transcendent" defined here (http://mathworld.wolfram.com/LerchTranscendent.html) but it doesn't seem to be any more of a useful form that what you give in the question.
The partial sums look like:

