Does the series $\sum_{n=0}^{\infty }\frac{1}{n+n^2x^2}$ converge uniformly on $x\in(a,\infty )\; a\geq 0$

Question

My Attempt: The sequence $$< \frac{1}{n+n^2x^2}>$$ converges uniformly on $x\in(a,\infty ) \; a\geq 0$ to $0$.

But I have little confusion about series $$\sum_{n=0}^{\infty }\frac{1}{n+n^2x^2}$$ How to show that series converges uniformly on $x\in(a,\infty )\; a\geq 0$?

Thanks in advance!

Answer 1

It is not uniformly convergent on $(0,\infty)$. Choose a sequence $n_k$ with $n_{k+1} >2n_k+1$. Consider $\sup _x \sum\limits_{n=n_k+1}^{n_{k+1}} \frac 1 {n+n^{2}x^{2}} >\sum\limits_{n=n_k+1}^{n_{k+1}} \frac 1 { n_{k+1}+1}$ as seen by taking $x=\frac 1{\sqrt {n_{k+1}}}$. Hence $\sup _x \sum\limits_{n=n_k+1}^{n_{k+1}} \frac 1 {n+n^{2}x^{2}} >\frac 1 2$ for all $k$. This proves that the series is not uniformly convergent on $(0,\infty)$.

It may be noted that if $a>0$ then for $x \geq a$ we have $\frac 1 {n+n^{2}x^{2}} \leq \frac 1 {n^{2}a^{2}}$. It follows that the series converges uniformly on $[a,\infty)$ for each $a >0$.