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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...

3 Answers 3

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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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    @PeterTamaroff: This follows from did's answer. Suppose $x$ is in some unbounded subset of $\mathbb{R}$ and that there is no $R$ such that $x\le R$. For every $k$ we can find an $x\ge k$. But then \sum_{n=k}^\infty\frac{x}{x^2+n^2} \geq\int_{k}^{+\infty}\frac{x}{x^2+t^2}\, dt \geq\int_{k}^{+\infty}\frac{k}{k^2+t^2}\, dt >0.2012-07-01
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There is no uniform convergence since, for every positive integer $x$, $\sum_{n\geqslant x}\frac{x}{x^2+n^2}\geqslant\int_x^\infty\frac{x}{x^2+t^2}\,\mathrm dt=\frac\pi4\gt0,$ 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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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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    oenamen: My previous comment was explicitely directed at Lag. (But your *very large subsets* made me smile.)2012-07-05