Uniform convergence of $\sqrt{x^2+n^2}^{-1}-n^{-1}$ 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 $a&gt;0$.

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 $\mathbb R.$
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 $a&gt;0$.

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 $\mathbb R.$