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 $x$?

  • $\begingroup$ 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

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 $$ \sum_{n\ge M}\frac{1}{(M-\varepsilon-n)^2}=\frac{1}{\varepsilon^2}+ \sum_{n\ge M+1}\frac{1}{(M-\varepsilon-n)^2}\le\frac{1}{\varepsilon^2}+ \sum_{n\ge 1}\frac{1}{n^2}. $$

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.