Uniform convergence of functional series $\sum\limits_{n=1}^{\infty}\frac{1}{x+n}*th\frac{x}{\sqrt{n}}$

Question

How to investigate the uniform convergence of the series $\sum\limits_{n=1}^{\infty}\frac{1}{x+n}*th\frac{x}{\sqrt{n}}$ on the $(1, +\infty)$?

I tried all the possible tests to prove it's convergence and didn't succeed. So i decided to prove it's uniform divergence by Cauchy’s Criterion: $\exists e>0 \forall k(e) \exists K\ge k(e), \exists p , \exists x: |\sum\limits_{n=k+1}^{k+p}\frac{1}{x+n}*th\frac{x}{\sqrt{n}}| \ge e$.

So I took $p=2k:$ $|\sum\limits_{n=k}^{2k}\frac{1}{x+n}*th\frac{x}{\sqrt{n}}| \ge...\ge e$? What's next (if my previous steps are correct of course)? Which $x=x(n)$ should I take to to get something that does not tend to zero?

Answer 1

I've done it:

$\exists e=0.01 : \forall k(e) \exists K=2 k(e), \exists p=K , \exists x=\sqrt{k}: |\sum\limits_{n=k}^{k+p}\frac{1}{x+n}*th\frac{x}{\sqrt{n}}| = |\sum\limits_{n=k}^{2k}\frac{1}{\sqrt{k}+n}*th\frac{\sqrt{k}}{\sqrt{n}}| \ge |\sum\limits_{n=k}^{2k}\frac{1}{\sqrt{k}+2k}*th\frac{\sqrt{k}}{\sqrt{2k}}| = |\frac{k}{\sqrt{k}+2k}*th\frac{1}{\sqrt{2}}| \ge |\frac{k}{2k+2k}*th\frac{1}{\sqrt{2}}|= \frac{1}{4}*th\frac{1}{\sqrt{2}} \ge0.01= e$

Thus we have a uniform divergence of this series.

Answer 2

$$ \frac{1}{x+n}\text{th}\left(\frac{x}{\sqrt{n}}\right) \leq \frac{x}{\left(x+n\right) \sqrt{n}} $$ Let $x \in \left[a,b\right] \subset \left]1,+\infty\right[$ $$ \left\|\frac{1}{x+n}\text{th}\left(\frac{x}{\sqrt{n}}\right)\right\|_{\infty,\left[a,b\right]} \leq \frac{b}{\sqrt{n}\left(n+1\right)} \underset{(+\infty)}{\sim}\frac{b}{n^{3/2}} $$ And the series $\displaystyle \sum_{n \geq 1}\frac{b}{n^{3/2}}$ converges so we have normal convergence on every compact set of $\left]1,+\infty\right[$ and hence uniformly ( on every compact set of $\left]1,+\infty\right[$ )