Determining whether the complex series $\sum_{n=0}^\infty \frac{n^2}{(n + i)(n + 200 + 2i)}$ converges I'd like to verify whether the following reasoning is valid for determining whether the complex series $$\sum_{n=0}^\infty \frac{n^2}{(n + i)(n + 200 + 2i)}$$ converges or not: With the $n$-th term divergence test we see that $$\frac{n^2}{(n + i)(n + 200 + 2i)} = \frac{n^2}{n^2(1 + i/n)(1 + (200/n) + (2i/n))} = \frac{1}{(1 + i/n)(1 + (200/n) + (2i/n))}$$ so that as $n \to \infty$, $$\frac{1}{(1 + i/n)(1 + (200/n) + (2i/n))} \to \frac{1}{(1 + 0)(1 + 0 + 0)} = 1 \neq 0$$ Thus the series diverges.
 A: Think about WHY the test you're using works, and it can give you some insight as to when you can and can't apply certain tests that you know. In this case you can apply this test, but the tests are slightly different in general. Why does $\lim z_n\neq 0$ imply $\sum_{n=0}^{\infty}$ diverges? On an intuitive level, say $\lim z_n=1$. Thus, for large $n$, we're basically adding $1+1+1+1+...$ forever, which obviously diverges.
How does this apply to your series when we move to $\mathbb{C}$? Well... we can just consider whether $z_n$ approaches some other complex number. It's just another case we have to consider. Say $\lim z_n = 1+i.$ Then, for large $n$, we'd essentially be adding $1+i+1+i+1+i+...$ forever, which ALSO diverges. See how we can use the exact same reasoning for complex series as we did for real series?
This was a very long-winded response, as it's much easier to say "Yep. You're right." But hopefully you can realize why the Test for Divergence can also be applied to complex series.
