Regularizing infinite sum over $\sqrt{n^2+a^2}$

Question

I am aware that one can use zeta function regularization to obtain \begin{equation} \sum_{n\in \mathbb{N}}n = -\frac{1}{12} \end{equation} I am wondering if it is possible to regularize a similar sum, namely \begin{equation} \sum_{n\in \mathbb{N}} \sqrt{n^2+a^2} \end{equation} I assume one needs to study a function F(a,s) which in some convergent region can be written as \begin{equation} F(a,s) = \sum_{n\in \mathbb{N}} \frac{1}{(n^2+a^2)^s} \end{equation} But I have no idea what this function is, or how it relates to other known functions. Any help is appreciated.

Answer 1

The usual disclaimers apply to the readers when talking about regularizations: I will be using the notation $\sum_{n=1}^\infty a_n$ to denote the regularization. This is not necessarily the same as the limit of the partial sums, and in most cases it's different. If you copy this answer into your calculus homework, it will be marked incorrect.

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Note that $$\sqrt{n^2+a^2} = \left(\sqrt{n^2+a^2}-n-\frac{a^2}{2n}\right) + n + \frac{a^2}{2n}$$ and $\sum_{n=1}^\infty \left(\sqrt{n^2+a^2}-n-\frac{a^2}{2n}\right)$ converges. Therefore, any regular, linear summation method that assigns a value to $\sum_{n=1}^\infty \sqrt{n^2+a^2}$ will assign a value to $\sum_{n=1}^\infty n$ if and only if it assigns a value to $\sum_{n=1}^\infty \tfrac1n$.

Assuming this, we have $$\sum_{n=1}^\infty \sqrt{n^2+a^2} = \sum_{n=1}^\infty\left(\sqrt{n^2+a^2}-n-\frac{a^2}{2n}\right) + \sum_{n=1}^\infty n + \frac{a^2}2\sum_{n=1}^\infty\frac1n.$$

Now for the standard zeta regularization, $\sum_{n=1}^\infty n = -1/12$, but $\sum_{n=1}^\infty\frac1n$ corresponds $\zeta(1)$, where there is a pole. Therefore, we need to extend zeta regularization. There's a couple standard ways to do this, e.g. via Cauchy principal values or Ramanujan summation, which give the same result: $\sum_{n=1}^\infty 1/n = \gamma$ (the Euler-Mascheroni constant).

To summarize, this shows that, in the natural extension of the zeta function regularization by either Cauchy principal values or Ramanujan summation, we have $$\sum_{n=1}^\infty \sqrt{n^2+a^2} = \sum_{n=1}^\infty \left(\sqrt{n^2+a^2}-n-\frac{a^2}{2n}\right) - \frac1{12}+\frac12a^2\gamma$$ where the only remaining summation on the right-hand side is an honest introductory calculus series which is the limit of partial sums.

What is that value in terms of $a$? I have no idea.

Answer 2

Nice example! Not every divergent series can be assigned a value, and it looks like you’ve stumbled upon one of those. To see why, let’s build up a few useful facts about the series involved. First, observe that we can write $\sqrt{n^{2}+a^{2}}=a\sqrt{\left(\frac{n}{a}\right)^{2}+1}$, so we only need to consider the Taylor series $\sqrt{x^2+1}$. The Taylor for this is given by $$\sqrt{x^2+1} = \sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)!}{\left(\frac{1}{2}-n\right)!n!}x^{2n}$$ Where, as usual, the factorial at non-integer values is interpreted in terms of the gamma function. What we really want though, is a Taylor expansion of the sqrt function about infinity. Unfortunately, such an expansion doesn’t exist, since the branch cut means $+\infty$ and $-\infty$ won’t mesh together in the right way. However, we can get a Taylor-like expansion by noting that $\sqrt{x^{2}+1}=\left|x\right|\sqrt{\left(\frac{1}{x}\right)^{2}+1}$ so then we have the expansion in the positive direction as $$\sum_{n=0}^{\infty}\frac{\left(\frac{1}{2}\right)!}{\left(\frac{1}{2}-n\right)!n!}x^{1-2n}$$ Since $(n/a)^2 > 0$, we use the expansion in the positive direction. Now, we are set up to look at the zeta-regularization of the original series. We have $$\sum_{n=1}^{\infty}\frac{\sqrt{n^{2}+a^{2}}}{n^{x}}=\sum_{n=1}^{\infty}\frac{1}{n^{x}}\sum_{k=0}^{\infty}a^{2k}\frac{\left(\frac{1}{2}\right)!}{\left(\frac{1}{2}-k\right)!k!}n^{1-2k}=\sum_{k=0}^{\infty}a^{2k}\frac{\left(\frac{1}{2}\right)!}{\left(\frac{1}{2}-k\right)!k!}\sum_{n=1}^{\infty}n^{\left(1-2k-x\right)}$$ $$F(x,a)= \sum_{k=0}^{\infty}a^{2k}\frac{\left(\frac{1}{2}\right)!}{\left(\frac{1}{2}-k\right)!k!}\zeta\left(2k+x-1\right)$$ For large enough $x$, all of these steps are valid, and so this provides us with an analytical continuation for $F(x,a)$ (though, note that it is only valid for $a \leq 1$). Now, we are able to evaluate, for instance, $$F(-1,1) = \sum_{n=1}^\infty n \sqrt{n^2+1}= -\frac{1}{4}-\frac{\pi^{2}}{48}+\frac{\pi^{4}}{1440}-\dots \approx -0.4118$$ However, the function has a pole at $x=2-2k, k \in \mathbb{Z}$, so this expansion suggests that the original series should diverge, since its expanded around a pole. However, we can find that it has a simple pole of $\frac{1}{z}\frac{a^2}{2}$. The other answer is basically analyzing the constant term in the expansion around $x=0$. So, we can finish off that analysis to obtain the missing terms. We have that $$\sum \sqrt{n^2+a^2} “=“ -\frac{1}{12} + \frac{\gamma}{2} a^2 + \sum_{k=2}^{\infty}a^{2k}\frac{\left(\frac{1}{2}\right)!}{\left(\frac{1}{2}-k\right)!k!}\zeta\left(2k-1\right)$$ Thus, this divergent series doesn’t have a well-defined value from the point of view of analytical continuation. However, we can extract out its constant term as the next best value to assign to this series.

Note that you can get an analytical continuation of $\sum_{n=0}^\infty \frac{1}{(n^2+a^2)^s}$ for general $s$ in an essentially identical way. We expand the inside using the binomial formula and then switch the order of summation to obtain something in terms of the zeta function.