Assume the following sum: $$ \sum_{n=1}^{\infty} f_n(x) = \sum_{n=1}^{\infty}\left(\frac{1}{\sqrt{x^2+n^2}}-\frac{1}{n}\right), $$ where $x \in \mathbb{R}$. The problem is to determine whether the sum converges uniformly, i.e. whether there exists $s(x)$ such as $\sum f_n(x) \rightrightarrows s(x)$. My approach is following: I show $f_n \rightarrow 0$ for every $x$, as $$ f_n \sim \frac{1}{n^3}, \lim_{n \rightarrow \infty} \frac{f_n}{n^{-3}} = -\frac{x^2}{2}. $$ Then I can show $f_n \rightrightarrows 0$, as $$ |f_n(x)| \leq \frac{1}{n}, \forall x, $$ thus the sequence $f_n$ converges uniformly to zero, and here I got stuck. Is my approach correct? Could you help me find out on which interval the sum converges uniformly? Is the convergence absolute/local? What is the prescription of $s(x)$?
Thank you for any advice.
For fixed $x$,
$$f_n(x)=\frac{1}{n}((1+(\frac{x}{n})^2)^{-\frac{1}{2}})-\frac{1}{n}$$
$$=\frac{1}{n}(1-\frac{1}{2}\frac{x^2}{n^2}(1+\epsilon(n)))-\frac{1}{n}$$
$$=-\frac{x^2}{2n^3}(1+\epsilon(n))$$
thus the series $\sum f_n(x)$ converges.
Now let $a>0$.
for large enough $n$
$\forall x\in [-a,a]$
$|f_n(x)|\leq \frac{a^2}{2n^3}$
thus the series converges uniformly at
$ [-a,a]$ and also at $[a,b]$ but not on $\mathbb R.$