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

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?

• For large $n$, $|n+z|$ if of the same ordes as $n$. This means that you can, for example, show that, for fixed $z$, and large $n$, $|n+z|^2 \ge C |n|^2$ (with some Constant $C = C(z)$) – Thomas Oct 23 '17 at 15:53
• If $R$ is large enough the domain $D$ encloses a double pole of the given function ($\psi'(z+1)$) and the convergence cannot be uniform. – Jack D'Aurizio Oct 23 '17 at 16:42
Hint. For $n>R$ and $|z|\leq R$, we have that $|n+z|\geq n-|z|\geq n-R>0$, and $$\left\vert\frac{1}{(n+z)^2}\right\vert \lt\frac{1}{(n-R)^2}.$$