What does $\sum\limits_{n=1}^{\infty}\left( {\frac{1}{n}-\frac{1}{\sqrt{n^2+x^2}}} \right)$ equal? I tried calculating it in Mathematica but it can't seem to figure out a formula for this. If you define x as a specific real number then the sum always converges and gives you a number. The output seems to be logarithmic because when I do $e^{\sum_{n=1}^{\infty}{\frac{1}{n}-\frac{1}{\sqrt{n^2+x^2}}}}$ the output becomes linear as the function goes to $\infty$. It isn't linear as x approaches 0 though. Does anyone know the answer to this or how to find it?
Here is a plot of $e^{\sum_{n=1}^{\infty}{\frac{1}{n}-\frac{1}{\sqrt{n^2+x^2}}}}$

 A: For $x$ sufficiently small we have that
$$\frac{1}{\sqrt{n^2+x^2}}=\frac 1 n\frac{1}{\sqrt{1+(x/n)^2}}\sim\frac1n\left(1-\frac12\frac{x^2}{n^2}\right)=\frac 1n-\frac12\frac{x^2}{n^3}$$
thus
$$\frac{1}{n}-\frac{1}{\sqrt{n^2+x^2}} \sim \frac12\frac{x^2}{n^3}$$
and therefore
$$S(x)=\sum_{n=1}^{\infty}\left({\frac{1}{n}-\frac{1}{\sqrt{n^2+x^2}}}\right)\sim \frac12x^2\sum_{n=1}^{\infty} \frac1{n^3}=cx^2$$
which explains qualitatively the nonlinear beaviour for $x$ sufficiently small.
For $x$ large let consider an integral estimation that is
$$S(x,k)=\sum_{n=1}^{k}\left({\frac{1}{n}-\frac{1}{\sqrt{n^2+x^2}}}\right)\approx \int_1^k \left({\frac{1}{n}-\frac{1}{\sqrt{n^2+x^2}}}\right)dn=\left[\log \frac{n}{\sqrt{n^2+x^2}+n}\right]_1^k=\log \frac{k}{\sqrt{k^2+x^2}+k}+\log (\sqrt{1+x^2}+1)\to -\log 2+\log (\sqrt{1+x^2}+1)\sim \log x$$
A: Interesting question. The inverse Laplace transform gives 
$$ \frac{1}{n}-\frac{1}{\sqrt{n^2+x^2}} = \int_{0}^{+\infty} (1-J_0(xs))e^{-ns}\,ds $$
hence
$$ \sum_{n\geq 1}\left(\frac{1}{n}-\frac{1}{\sqrt{n^2+x^2}}\right) = \int_{0}^{+\infty}\frac{1-J_0(xs)}{e^s-1}\,ds $$
where the Bessel function $J_0(t)$ behaves like $1-\frac{t^2}{4}$ in a right neighbourhood of the origin and like $\frac{\sin(t)+\cos(t)}{\sqrt{\pi t}}$ in a left neighbourhood of $+\infty$ (Tricomi's inequality). This allows a simple numerical approximation of the wanted series. If $x$ is close to zero we have
$$ \sum_{n\geq 1}\left(\frac{1}{n}-\frac{1}{\sqrt{n^2+x^2}}\right) \approx x^2\frac{\zeta(3)}{2}-x^4\frac{3\zeta(5)}{8}+\ldots $$
