# 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I'm assuming this sum cannot be evaluated exactly (although, the book is old from 1975, maybe it's out of date?)...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?

• Of course $F(x) \to 0$ as $x \to 0$ (by monotone convergence thm for example) – mathworker21 Jan 18 '18 at 7:31
• Without better esteimates, $\sum\mathcal O({x^4})\ne \mathcal O({x^2})$ (or anything similar) – Hagen von Eitzen Jan 18 '18 at 7:39
• Your answer is right. It can be justified using the following facts: $e^{-t} \geq 1-t$ and $e^{-t} \leq 1-t+\frac {t^{2}} 2$ and $\frac {ln(1+x)} x \to 1$ as $x \to 0$. – Kabo Murphy Jan 18 '18 at 8:10
• Or you can use the identity $$F(x) = \sum_{n=1}^{\infty} \left( \frac{1}{n} - \frac{1}{\sqrt{n^2+x^2}} \right) = \int_{0}^{x} \sum_{n=1}^{\infty} \frac{t}{(n^2+t^2)^{3/2}} \, dt$$ to show that $$\forall x\in\mathbb{R} \ : \quad |F(x)| \leq \frac{\zeta(3)}{2}x^2.$$ – Sangchul Lee Jan 18 '18 at 8:15

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 $J_0(sx)\approx \frac{\sin(sx)+\cos(sx)}{\sqrt{\pi s x}}$.

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 $(2)$ $$\lim\limits_{x\rightarrow 0+}F(x)=0$$

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

$$\sum_{n=1}^{\infty} \left[ \frac{1}{n} - \frac{1}{\sqrt{ n^2 + x^2 }} \right]=\sum_{n=1}^{\infty} \left[ \frac{1}{n} - \frac1n+\frac{x^2}{2n^3}+o\left(\frac{1}{n^3}\right) \right]=\sum_{n=1}^{\infty} \left[ \frac{x^2}{2n^3}+o\left(\frac{1}{n^3}\right) \right]$$

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 $\frac{\zeta(3)}2\,x^2$.