Let $f_k:(0,\pi) \to \mathbb{R}$ be a sequence such that $f_k(x) = \displaystyle \sum_{k=1}^{\infty} \frac{2x}{x^2 - (\pi k)^2}$, show that $f_k$ converges uniformly.
I need to confirm if it's correct, can you read my solution?
My solution:
I wanna show by Uniform Cauchy Criterion, that is, $\forall \epsilon>0,\exists n_0 \in \mathbb{N}$, such that $m,n >n_0 \Rightarrow |f_m(x) - f_n(x)|< \epsilon, \forall x\in (0,\pi).$
Ok, so let $\epsilon>0$ and take $n_0> \sqrt{\frac{2}{\epsilon}+1}$, then for $m>n>n_0$ we have
$|f_m(x)- f_n(x)| = \left|\displaystyle \sum_{k=n+1}^{m} \frac{2x}{x^2 - (\pi k)^2} \right| \leq \displaystyle \sum_{k=n+1}^{m} \left| \frac{2x}{x^2 - (\pi k)^2} \right| \leq \frac{1}{\pi} \displaystyle \sum_{k=n+1}^{m} \frac{2}{k^2-1}.$
The first inequality is given by Triangle inequality and the last one it's because $x \in (0,\pi)$ and that implies that $2x < 2\pi$, $\left|\frac{1}{x^2 - (\pi k)^2}\right| < \left|\frac{1}{\pi^2 - (\pi k)^2}\right|$ and $|1-k^2|= k^2-1$.
Now, since $m>n>n_0 > \sqrt{\frac{2}{\epsilon}+1}$ we have
$\frac{1}{\pi} \displaystyle \sum_{k=n+1}^{m} \frac{2}{k^2-1} < \frac{1}{\pi} \displaystyle \sum_{k=n+1}^{m} \epsilon = \frac{(m-n)\epsilon}{\pi}.$
It's ok? the difference $m-n$ is very weird, should I be careful about this?
That is not correct, as $m,n$ are still running, the term $(m-n)\epsilon/\pi$ can tend to infinity, so the inequality is not being controlled by arbitrary small.
A way to correct that is to control $\displaystyle\sum_{k=n+1}^{m}\dfrac{2}{k^{2}-1}$, we know that $\displaystyle\sum_{k=1}^{\infty}\dfrac{2}{k^{2}-1}<\infty$, so by Cauchy criterion to the series, you may choose an $n_{0}'$ such that $\displaystyle\sum_{k=n+1}^{m}\dfrac{2}{k^{2}-1}<\epsilon$ for all $m>n\geq n_{0}'$. Now you take $\max\{n_{0},n_{0}'\}$.