I'm learning about series of functions and need some help with this problem :

Given the series of functions $$ \sum_{n=1}^\infty \frac{x}{x^2+n^2}, \; x \in (0, \infty) $$ show that it converges pointwise and uniformly to a function $ s(x) $ and show that $ s'(x) $ converges uniformly.

Here's my attempt :

Let $$ f_n(x) = \frac{x}{x^2+n^2} \Rightarrow f_n'(x) = \frac{(x^2 + n^2) - 2x^2}{(x^2+n^2)^2} = \frac{n^2 - x^2}{(x^2+n^2)^2} $$


$$ f_n'(x) = 0 \iff n^2 - x^2 = 0 \iff x = \pm \, n $$

On the interval $ (0, \infty), \; f_n $ takes it's maximum value for $ x = n $ and therefore :

$$ f(n) = \frac{n}{n^2+n^2} = \frac{1}{2n} $$

Hence $ |f_n(x)| \le \frac{1}{2n} $ but the series $ \sum_{n=1}^\infty \frac{1}{2n} $ does not converge by the p-series Test.

What am I doing wrong here?

  • 1
    $\begingroup$ If $|f_n(x)|\le M_n$ for all $x$ and $\sum M_n<\infty$ then $\sum f_n$ converges uniformly. But that only works in one direction, it's not an if and only if thing. It's possible for $\sum f_n$ to converge uniformly even though $\sum M_n=\infty$. So you haven't necessarily done anything wrong, but you still have to show that series converges uniformly by some other method... $\endgroup$ – David C. Ullrich Nov 8 '15 at 23:52
  • $\begingroup$ Hint: For each $x$, consider the terms with $n\le x$ and the terms with $n>x$ separately. For $n\le x$ use the fact that $x/(x^2+n^2)\le x/x^2$; for $n>x$ use the fact that $x/(x^2+n^2)\le x/n^2$. $\endgroup$ – David C. Ullrich Nov 8 '15 at 23:56
  • $\begingroup$ As an aside, $\displaystyle\sum_{n=-\infty}^\infty\frac x{x^2+n^2}~=~\pi~\coth\pi x.$ This can be shown by differentiating the natural logarithm of Euler's infinite product formula for the sine function. $\endgroup$ – Lucian Nov 9 '15 at 8:15
  • $\begingroup$ @DavidC.Ullrich: In fact, that line of reasoning proves that it doesn't converge uniformely. $\endgroup$ – Eric Naslund Dec 9 '15 at 17:33
  • $\begingroup$ @EricNaslund Eewps... $\endgroup$ – David C. Ullrich Dec 9 '15 at 18:07

This series does not converge uniformely.

Proof: Suppose that it did converge uniformely. Then for any $\epsilon>0$ there exists $N$ such that $$\left|\sum_{n>N}\frac{x}{x^{2}+n^{2}}\right|<\epsilon$$ uniformely for all $x\in(0,\infty)$. Let $\epsilon=1/10$, let $N$ be given and set $x=N$. Then $$\left|\sum_{n>N}\frac{x}{x^{2}+n^{2}}\right|>\sum_{N<n\leq2N}\frac{N}{N^{2}+n^{2}}\geq\sum_{N<n\leq2N}\frac{N}{5N^{2}}=\frac{1}{5}>\epsilon.$$


Aside from using the $\text{ Cauchy Criterion for Series of Functions }$ let's let the following :

$$F(x)=\lim_{n\to\infty}F_n(x)=\lim_{n\to\infty}\sum_{j=1}^n{x\over x^2+j^2}=\lim_{n\to\infty}\sum_{j=1}^nf_j(x)\quad j\le n,\:x>0.$$

Now, $$\forall\varepsilon>0\:\:\forall x>0\:\:\exists N_{\varepsilon,x}\in\mathbf N\:\:\:\forall n\ge N_{\varepsilon,x}\implies |f_n(x)-f(x)|<\varepsilon\:.$$

Indeed, whenever $\:n\ge\:N_{\varepsilon,x}>\sqrt{{\Large\frac{x}{\varepsilon}}-x^2}\quad\&\quad \underbrace{1>\varepsilon x}_{\text{Archimedes}},\:\:$ we have that $\:|f_n(x)-\underbrace{f(x)}_{=\:0}|<\varepsilon$.

Thus $\:f_n\:$ converges pointwise to $\:f\equiv0.$

But $$\lim_{n\to\infty}\left[\lim_{T\to \infty}\int_0^Tf_n(x)\text{d}x\right]=\lim_{n\to\infty}\left[\lim_{T\to\infty}\frac{1}{2}\ln\left(1+\frac{T^2}{n^2}\right) \right]=\lim_{n\to\infty}\infty=\infty\\>\lim_{T\to \infty}\int_0^T\lim_{n\to\infty}f_n(x)\text{d}x=0.$$ So, $\:\underset{n\ge N}{f_n}\:$ is not uniformly convergent on $\:\mathbf R_+\implies \underset{n\ge N}{F_n} \:$ is not uniformly convergent on $\:\mathbf R_+$.


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