Proof of Convergence of Series?

Question

I am trying to show that the infinite sum (for a fixed complex number $z$) $$\sum_{n=1}^\infty\frac{z}{(z-n)n}$$ converges absolutely. My idea is to compare it to the series $$1+1/4+1/9+\cdots,$$ but for some reason I cannot seem to figure out how to handle the asymptotics. It seems like "in the long run" the terms of the series get closer together, so I thought of proving a general result like that, but then I realized that the harmonic series and $0+0+0+\cdots$ fits the same property.

Could someone supply the technical details?

Answer 1

Since$$\lim_{n\to\infty}\frac{\left|\frac z{(z-n)n}\right|}{\frac1{n^2}}=\lim_{n\to\infty}\left|\frac{n^2z}{nz-n^2}\right|=|z|,$$then, since the series $\sum_{n=1}^\infty\frac1{n^2}$ converges, your series converges absolutely.

Answer 2

You can decompose the terms $\frac{z}{(z-n)n}$ as $\frac{1}{z-n} + \frac{1}{n}$. So it becomes $$\sum_{n = 1}^{\infty} \frac{1}{z-n} + \frac{1}{n}$$

For $z=0$, it can be seen that series converges to 0 and for any $z \in \mathbb{Z}^+$ the series diverges. For other values of fixed $z$, it can be observed that these terms cancel each other out in the tail (when $n \gg z$).