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 $e^{\sum_{n=1}^{\infty}{\frac{1}{n}-\frac{1}{\sqrt{n^2+x^2}}}}$


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
$$S(x,k)=\sum_{n=1}^{k}\left({\frac{1}{n}-\frac{1}{\sqrt{n^2+x^2}}}\right)\approx \int_1^k \left({\frac{1}{n}-\frac{1}{\sqrt{n^2+x^2}}}\right)dn=\left[\log \frac{n}{\sqrt{n^2+x^2}+n}\right]_1^k=\log \frac{k}{\sqrt{k^2+x^2}+k}+\log (\sqrt{1+x^2}+1)\to -\log 2+\log (\sqrt{1+x^2}+1)\sim \log x$$
