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Let $M>0$. $$\sum_{n=M}^{\infty} \frac {1}{(x-n)^2}$$

How do I show this converges uniformly for $x ≤ \frac{|M|}{2}$

My actual question is how do I determine the interval of uniform convergence for general series like this one.

Am I allowed to use the Weierstrass M test? Because the sum is between $M$ and $\infty$.

$\sum_{n=M}^{\infty} \frac {1}{(x-n)^2}≤\sum_{n=M}^{\infty} \frac {1}{n^2}$ which converges so the series converges uniformly for all $x$?

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  • $\begingroup$ What is $x$? If $x \in \mathbb Z$ and $x \geq M$ then there will be a problem... $\endgroup$ – Théophile Nov 24 '15 at 0:14
  • $\begingroup$ Edited my question $\endgroup$ – Jack Nov 24 '15 at 0:23
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In fact, this series converges uniformly in $(-\infty,M-\varepsilon]$, for every $\varepsilon>0$, since if $x\in(-\infty,M-\varepsilon]$, then $$ 0<\frac{1}{(x-n)^2}\le \frac{1}{(M-\varepsilon-n)^2},\quad\text{for all $n\ge M$}, $$ and hence $$ 0<\sum_{n\ge M}\frac{1}{(x-n)^2}\le \sum_{n\ge M}\frac{1}{(M-\varepsilon-n)^2},\quad\text{for all $n\ge M$}. $$ Hence, it suffices to show that the series $\sum_{n\ge M}\frac{1}{(M-\varepsilon-n)^2}$ converges. To do this observe that $$ \sum_{n\ge M}\frac{1}{(M-\varepsilon-n)^2}=\frac{1}{\varepsilon^2}+ \sum_{n\ge M+1}\frac{1}{(M-\varepsilon-n)^2}\le\frac{1}{\varepsilon^2}+ \sum_{n\ge 1}\frac{1}{n^2}. $$

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  • $\begingroup$ How do I show it converges uniformly for x≤|M|/2? $\endgroup$ – Jack Nov 24 '15 at 1:08
  • $\begingroup$ Let epsilon be M/2 in your answer? $\endgroup$ – Jack Nov 24 '15 at 1:16

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