Absolute convergence of $\sum\limits_{n=0}^{\infty} \frac {z} {(z+n)^2}$

Question

I want to check the absolute convergence of $\displaystyle\sum_{n=0}^{\infty} \frac {z} {(z+n)^2}$ in the half plane $\Re(z)>0$, and to see if the convergence is uniform or locally uniform. How do I find the sum function?

Answer 1

To check the absolute convergence, use $$ |z+n|^2\geq |Re(z+n)|^2=(Re z + n)^2 $$ and comparison with $1/n^2$.

Because of $z$ in the numerator, the convergence of this series is locally uniform in any compact subset of the half plane $Re z > 0$.

I don't have the sum function for your summation, but there is a formula which gives the following summation. $$ \sum_{n=-\infty}^{\infty} \frac{1}{(z+n)^2}=\frac{\pi^2}{(\sin\pi z)^2},\ $$ for non-integral $z$. You can obtain this formula by contour integration of $$f(s)=\frac{\pi\cot\pi s}{(z+s)^2}$$