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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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    I agree with Peter, because for x=n, its: $\sum 1/n$2012-06-30
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    This looks okay to me. (+1) @Lag: We can't set $x=n$, $n$ ranges over $1,2,\ldots$.2012-06-30
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    @oenamen The OP asks for every $x$. I think we should be thinking about the *whole* real line.2012-06-30
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    @PeterTamaroff: $\infty$ is not a real number.2012-06-30
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    @oenamen [This](http://en.wikipedia.org/wiki/Extended_real_number_line).2012-06-30
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    @PeterTamaroff: I see what you're getting at after reading did's answer. For uniform convergence we need to restrict ourselves to a bounded subset of $\mathbb{R}$.2012-06-30
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    @oenamen That is what needs to be proven then: convergence is uniform only in a bounded subset of $\Bbb R$.2012-06-30
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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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    That sounds really good. I think this might solved it.2012-06-30
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    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.2012-07-05
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    @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.2012-07-05
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    oenamen: My previous comment was explicitely directed at Lag. (But your *very large subsets* made me smile.)2012-07-05