<s>[INST] 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I'm assuming this sum cannot be evaluated exactly (although, the book is old from 1975, maybe it's out of date?)...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?
 [/INST] 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
$$\sum_{n=1}^{\infty} \left[ \frac{1}{n} - \frac{1}{\sqrt{ n^2 + x^2 }} \right]=\sum_{n=1}^{\infty} \left[ \frac{1}{n} - \frac1n+\frac{x^2}{2n^3}+o\left(\frac{1}{n^3}\right) \right]=\sum_{n=1}^{\infty} \left[ \frac{x^2}{2n^3}+o\left(\frac{1}{n^3}\right) \right]$$
</s>