Consider the (complex) power series $$\sum_{n=1}^\infty \frac{(-1)^n}{n}z^{n(n+1)}$$
Show that the radius of convergence is $1$ and examine convergence for $z = 1,-1,i$.
Source problem: Conway's "functions of one complex variable".
Attempt:
For $z=1$, we get $\sum_{n=1}^\infty (-1)^n/n$ which converges by the alternating series test.
For $z=-1$, we get $$\sum_{n=1}^ \infty \frac{(-1)^{n+n(n+1)}}{n}= \sum_{n=1}^\infty \frac{(-1)^{n^2}}{n}= \sum_{n=1}^\infty \frac{(-1)^n}{n}$$ which is the same series as the previous case.
For $z=i$, we get
$$\sum_{n=1}^\infty \frac{(-1)^n}{n} i^{n(n+1)}$$
How can I deduce if this series converges/diverges? I tried ratio and root test but they both give limit $1$, so they do not apply. What other test can I use or how can I manipulate this expression in something I can work with?
If it follows that we have divergence for this value $z=i$, then it will also automatically follow that the convergence radius is $1$.
The sequence $\bigl((-1)^ni^{n(n+1)}\bigr)_{n\in\mathbb N}$ is simply the sequence$$1, -1, -1, 1, 1, -1, -1, 1,\ldots$$Since the partial sums of this sequence are bounded, and since the sequence $\left(\frac1n\right)_{n\in\mathbb N}$ is monotonic and converges to $0$, your series converges, by Dirichlet's test.