Check uniform convergence of $\sum_{n=1}^{\infty}\frac{x\sqrt{n}}{1+x^2n^2}$

Question

I need to check if $\sum_{n=1}^{\infty}\frac{x\sqrt{n}}{1+x^2n^2}$ for $x>0$ converges uniformly. I solved easier version of this problem which was to check if $\sum_{n=1}^{\infty}\frac{\sqrt{n}}{1+x^2n^2}$ converges uniformly for $x>0$, and I proved that it does not converge uniformly using Cauchy criterion. However I don't know how to prove the other one. I bounded it from above by something nicer: $$\sum_{n=1}^{\infty}\frac{x\sqrt{n}}{1+x^2n^2} < \sum_{n=1}^{\infty}\frac{1}{xn^{\frac{3}{2}}}$$ but I don't think that I can use this to prove uniform convergence.

Could anyone please help me solve this problem?

Answer 1

Let $f_n(x)=\dfrac{x\sqrt n}{1+x^2n^2}\in\mathcal C^1(\mathbb R)$.
Observe that $f_n\sim\dfrac{x\sqrt n}{x^2n^2}=\dfrac{1}{xn^{3/2}}$, so the set of punctual convergence is $\mathbb R$. $f'_n(x)=\dfrac{\sqrt n-x^2n^{5/2}}{(1+x^2n^2)^2}$ so $f'_n(x)>0\iff 1-n^2x^2>0\iff|x|<\dfrac{1}{n}$. It follows that $$\sup_{x\in D}|f_n(x)|=f_n(1/n)=\dfrac{1}{n}\dfrac{\sqrt n}{2}=\dfrac{1}{2\sqrt n}$$ so we can't have total convergence in a neighbourhood of $0$. In particular, when we study $x$ near zero, say $|x|\leq1/N$, we have $$\sup_{x\in \mathcal U(0)}\left\vert \sum_{n=N+1}^{2N}\dfrac{x\sqrt n}{1+x^2n^2}\right\vert\ge\sum_{n=N+1}^{2N}\dfrac{1/N\sqrt n}{1+n^2/N^2}\ge N\dfrac{\sqrt{N}}{N}\dfrac{1}{1+\frac{(2N)^2}{N^2}}$$ hence we have to exclude a neighbourhood of zero from the uniform convergence set. However, if you consider a point $x\in[K,+\infty)$, for $K>0$, you know that $f_n$ are decreasing in $[K,+\infty)$ then $$\sup_ {x\in[K,+\infty)}|f_n(x)|\le \dfrac{K\sqrt n}{1+K^2n^2}.$$ Since $f_n(-x)=-f_n(x)$, we have uniform convergence in $D_K=(-\infty,-K]\cup[K,+\infty)$, $K>0.$

Answer 2

We can consider an even "worse" series (at least in a finite interval $(0,M]$) with a general term $$a_n(x)=\frac{x}{1+n^2x^2}.$$

Consider that $\sum^\infty_{n=P} a_n \approx x \int_P^\infty \frac{dt}{1+x^2t^2} = x \left(\frac{(\pi/2) -\arctan(Px)}{x}\right) = \pi/2 - \arctan(Px)$. Now just set for example $x = \frac{1}{P}$ to show that there is no uniform convergence.

The $\approx$ sign is justified by the fact that $a_n(x)$ is uniformly decreasing in $n$.

Finally you can reduce your problem to this case by noticing that $\sqrt n>1$ in the Cauchy crtierion in order to prove that the series is ineed not uniformly convergent in in any interval $(0, M]$.