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How to prove that $\sum\limits_{n=1}^\infty \frac{x}{x^2+n^2}$ is uniformly convergent for every $x$?

I was trying all sort of ways, but it think the answer might be in solving the problem for $|x|<1$ and then for $|x|>1$.

Its easy to show that for $|x|<1$ , $\sum\limits_{n=1}^\infty \frac{x}{x^2+n^2} \leq \sum\limits_{n=1}^\infty \frac{1}{n^2}$ and using Weierstrass M-Test that the sum is uniformly convergent.

But for $|x|>1$ its a different story.

Does any one have a simple solution? I'm stuck...

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up vote 1 down vote accepted

We assume $x$ is in some bounded subset of $\mathbb{R}$, i.e., $|x|\le R$ for some real $R$.

Then $$\begin{equation*} \left|\frac{x}{x^2+n^2}\right| \le \frac{R}{x^2+n^2} \le \frac{R}{n^2} \end{equation*}$$ But the sum $\sum_{n=1}^\infty 1/n^2$ is convergent. Thus, the series converges uniformly on any bounded subset of $\mathbb{R}$ by the Weierstrass M-test.

Addendum: The series converges uniformly only on bounded subsets and not on $\mathbb{R}$ as @did shows in his answer.

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That sounds really good. I think this might solved it. – Gal Jun 30 '12 at 23:02
Lag: If the question is as written in your post (uniform convergence or not) and if you answer the above (uniform convergence on bounded subsets), somebody might consider that you simply do not answer the question. – Did Jul 5 '12 at 7:55
@did: I agree that as the question is written, the answer is "no" as you found and will edit my answer to that effect. I also think it is reasonable to recognize that this series converges uniformly on very large subsets of the reals. – user26872 Jul 5 '12 at 8:23
oenamen: My previous comment was explicitely directed at Lag. (But your very large subsets made me smile.) – Did Jul 5 '12 at 8:31

There is no uniform convergence since, for every $k\geqslant1$, $$\sum_{n\geqslant k}\frac{k}{k^2+n^2}\geqslant\int_{k}^{+\infty}\frac{k}{k^2+t^2}\,\mathrm dt=\frac\pi4\gt0,$$ hence $$ \lim\limits_{k\to\infty}\left(\sup\limits_{x\geqslant0}\sum_{n\geqslant k}\frac{x}{x^2+n^2}\right)\ne0. $$

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$$\sum_{n} \dfrac{|x|}{x^2 + n^2} = |x| \sum_{n} \dfrac{1}{x^2 + n^2} < |x| \sum_{n} \dfrac{1}{n^2} < \infty.$$ for any x.

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I agree with Peter, because for x=n, its: $\sum 1/n$ – Gal Jun 30 '12 at 22:28
This looks okay to me. (+1) @Lag: We can't set $x=n$, $n$ ranges over $1,2,\ldots$. – user26872 Jun 30 '12 at 22:31
@oenamen The OP asks for every $x$. I think we should be thinking about the whole real line. – Pedro Tamaroff Jun 30 '12 at 22:32
@PeterTamaroff: $\infty$ is not a real number. – user26872 Jun 30 '12 at 22:32
@oenamen This. – Pedro Tamaroff Jun 30 '12 at 22:34

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