Uniform convergence of $\sum_{n=1}^{\infty}\frac{1}{(n+z)^2}$

Question

I need to check uniform convergence of the series $$\sum_{n=1}^{\infty}\frac{1}{(n+z)^2} $$ on $D$: $\vert z\vert \leq R \lt \infty$

I've tried to use Weierstrass M-test, but I can't find such convergent series $\sum_{n=1}^{\infty}a_n$ that $\vert\frac{1}{(n+z)^2}\vert \lt a_n$ for all $n$. Do you have any ideas?

Or maybe I can use something else instead of Weierstrass M-test?

Thanks for your help.

Answer 1

Hint. For $n>R$ and $|z|\leq R$, we have that $|n+z|\geq n-|z|\geq n-R>0$, and $$\left\vert\frac{1}{(n+z)^2}\right\vert \lt\frac{1}{(n-R)^2}.$$