What can be said about the series $\sum_{n=1}^{\infty} \left[ \frac{1}{n} - \frac{1}{\sqrt{ n^2 + x^2 }} \right]$ This is a sequel to this question.
I recently was browsing through Hansen's "A Table of Series and Products", and I miraculously found the sum that I was looking for:
$$
\sum_{n=1}^\infty K_{0}\left( n z \right) \ = \ \frac{\pi}{2 z} + \frac{1}{2} \log\left( \frac{z}{4 \pi} \right) + \frac{\gamma}{2} - \frac{1}{2} \sum_{n=1}^{\infty} \left[ \frac{1}{n} - \frac{1}{\sqrt{ n^2 + \frac{z^2}{4\pi^2} }} \right]
$$
What a beautiful sum (the entire book is full of such amazing results). Here, $\gamma$ is the Euler-Mascheroni constant and $K$ is the modified Bessel function of the second kind (of order 0).
I'm of course curious about the latter sum now:
$$
F(x) := \sum_{n=1}^{\infty} \left[ \frac{1}{n} - \frac{1}{\sqrt{ n^2 + x^2 }} \right]
$$
I'm assuming this sum cannot be evaluated exactly (although, the book is old from 1975, maybe it's out of date?)...
I am most curious, how does this function $F(x)$ look like in the limit $x \to 0$? Does it diverge? My guess would be yes since $x$ is in a denominator. My only thought of how to attack this series in the limit $x\to 0$ is to take the following series expansion in this limit:
$$
\frac{1}{\sqrt{n^2 + x^2}} = \frac{1}{n} + \frac{x^2}{2n^3} + \mathcal{O}\left( x^4 \right)
$$
Then in this limit we have something like;
$$
F[x] \approx \sum_{n=1}^{\infty} \left[ \frac{1}{n} - \frac{1}{n} - \frac{x^2}{2 n^3} \right] = \frac{1}{2} x^2 \zeta(3)
$$
...in terms of the Riemann-zeta function. This seems like an answer, but I am worried there is something wrong with taking an expansion in the argument of the sum. Is this okay? Is there a better approximation?
 A: The approximation you mentioned is just fine. In a very simple (or elementary) way we have 
$$\sqrt{n^2+x^2}\geq n \Rightarrow 0\leq \frac{1}{n} - \frac{1}{\sqrt{n^2+x^2}}$$
so (from partials sums an taking limits)
$$F(x) \geq 0, \forall x \tag{1}$$
Next thing to observe is
$$\frac{1}{n} - \frac{1}{\sqrt{n^2+x^2}}=\frac{\sqrt{n^2+x^2}-n}{n\sqrt{n^2+x^2}}\leq\frac{\sqrt{n^2+x^2}-n}{n^2}=\\
\frac{n^2+x^2-n^2}{n^2(\sqrt{n^2+x^2}+n)}\leq \frac{x^2}{2n^3}$$
and this leads to
$$0 \leq F(x) \leq \frac{x^2}{2}\sum\frac{1}{n^3}=\frac{x^2 \zeta(3)}{2} \tag{2}$$
applying squeeze theorem to $(2)$ 
$$\lim\limits_{x\rightarrow 0+}F(x)=0$$
A: Yes it is correct, note that
$$\frac{1}{\sqrt{ n^2 + x^2 }}=-\frac1n\left(1+\frac{x^2}{n^2}\right)^{-\frac12}=\frac1n\left(1-\frac{x^2}{2n^2}+o\left(\frac{1}{n^2}\right)\right)=\frac1n-\frac{x^2}{2n^3}+o\left(\frac{1}{n^3}\right)$$
thus
$$\sum_{n=1}^{\infty} \left[ \frac{1}{n} - \frac{1}{\sqrt{ n^2 + x^2 }} \right]=\sum_{n=1}^{\infty} \left[ \frac{1}{n} - \frac1n+\frac{x^2}{2n^3}+o\left(\frac{1}{n^3}\right) \right]=\sum_{n=1}^{\infty} \left[ \frac{x^2}{2n^3}+o\left(\frac{1}{n^3}\right) \right]$$
A: There is nothing miraculous here, the first identity comes from computing the inverse Laplace transform of $\frac{1}{\sqrt{n^2+x^2}}$, which is given by $J_0(sx)$. It follows that
$$ \sum_{n\geq 1}\left[\frac{1}{n}-\frac{1}{\sqrt{n^2+x^2}}\right] = \int_{0}^{+\infty}\sum_{n\geq 1}\left(1-J_0(sx)\right)  e^{-ns}\,ds = \int_{0}^{+\infty}\frac{1-J_0(sx)}{e^s-1}\,ds $$
and the LHS is $\leq \frac{x^2}{2}\zeta(3)$ since $1-J_0(sx)\leq \frac{x^2 s^2}{4}$ for $sx\geq 0$. For large values of $sx$ we have Tricomi's approximation $J_0(sx)\approx \frac{\sin(sx)+\cos(sx)}{\sqrt{\pi s x}}$.
A: Since
$$
(1-4x)^{-1/2}=\sum_{k=0}^\infty\binom{2k}{k}x^k
$$
we get
$$
\begin{align}
\sum_{n=1}^\infty\left[\frac1n-\frac1{\sqrt{n^2+x^2}}\right]
&=\sum_{n=1}^\infty\frac1n\left[1-\frac1{\sqrt{1+\frac{x^2}{n^2}}}\right]\\
&=\sum_{k=1}^\infty(-1)^{k-1}\binom{2k}{k}\frac{\zeta(2k+1)}{4^k}x^{2k}
\end{align}
$$
the first term of which is $\frac{\zeta(3)}2\,x^2$.
