## A sum I can only imagine being evaluated using the residue theorem

The challenge this time is to evaluate the following sum:

$$\sum_{n=1}^{\infty} \frac{1}{n^3\sin{\left (\sqrt{2} \, n\pi \right)}}$$

NB (20 Nov 2016) I just got word that Wolfram fixed the problem described below. See below for details.

It should be clear that the sum converges…right? No? Then how do we show this?

Numerical experiments are more or less helpful, but as one might expect, there is a bit of jumping around about some value. So, even though the OP stated that he had proven convergence, I just want to illustrate how convergence is achieved.

At issue is the factor $\sin{\sqrt{2} \pi n}$ of each term in the sum: when is this sine term dangerously close to zero? If we think about it for a bit, the worst-case scenario is when $2 n^2$ is one less or more than a perfect square. (Recall that $2 n^2$ can never be a perfect square.) That is, when

$$2 n^2 = m^2 \pm 1$$

for some $m \in \mathbb{N}$. In this case,

$$\sin{\sqrt{2} \pi n} = \sin{\left (\sqrt{m^2 \pm 1} \pi \right )}$$

For $n$ sufficiently large, i.e., $m$ large as well, we have

$$\sin{\sqrt{2} \pi n} \approx \sin{\left [m \pi \left (1 \pm \frac1{2 m^2} \right ) \right ]} = (-1)^m \sin{\frac{\pi}{2 m}} \approx (-1)^m \frac{\pi}{2 m}$$

Thus,

$$\left | \frac1{n^3 \sin{\sqrt{2} \pi n}} \right | \le \frac1{n^3 \frac{\pi}{2 \sqrt{2} n}} = \frac{2 \sqrt{2}}{\pi n^2}$$

and, because the worst-case scenario term is bounded by something times $1/n^2$, the series converges by comparison with the sum of $1/n^2$.

OK, now that we have established convergence, we may now evaluate the sum. I would like to use the residue theorem, but the sum doesn’t exactly fit the typical pattern of sums that are residue-friendly. Some measure of creativity will be necessary. In this case, note that, when we have introduced a factor of $\csc{\pi z}$, the residue at $z=n \in \mathbb{Z}$ has a factor of $(-1)^n$, which this sum does not have. To blunt this factor, we may introduce a term $\csc{(\sqrt{2}-1) \pi z}$ which, at $z=n$ is equal to $(-1)^n \csc{\sqrt{2} \pi n}$.

For this reason, we consider the contour integral

$$\oint_{C_N} \frac{dz}{z^3 \sin{\pi z} \, \sin{\left [\left ( \sqrt{2}-1 \right ) \pi z \right ]}}$$

where $C_N$ is the square having vertices $\pm \left (N-\frac12 \right )(1 \pm i)$. You can show that, as $N \to \infty$, the contour integral over $C_N$ goes to zero.

However, the contour integral has poles at the integers and at the integers times $\sqrt{2}+1$. The residue at $z=0$ may be evaluated by expansion in a Laurent series, as the pole here is of order $5$. This expansion looks like

$$\frac1{z^3} \frac1{\pi z \left (1-\frac{\pi^2 z^2}{3!} + \frac{\pi^4 z^4}{5!}+\cdots \right )} \frac1{(\sqrt{2}-1)\pi z \left (1-\frac{(\sqrt{2}-1)^2\pi^2 z^2}{3!} + \frac{(\sqrt{2}-1)^4 \pi^4 z^4}{5!}+\cdots \right )}$$

The coefficient of $1/z$ in this expansion is essentially the coefficient of $z^4$ in the expansion of the sine terms in parentheses, or

$$\frac{13 \pi^2}{90} \left ( \sqrt{2}-1 \right )$$

The residue at each pole $z=n \ne 0$ is simply

$$\frac{(-1)^n}{\pi n^3 \sin{(\sqrt{2}-1) \pi n}} = \frac1{\pi n^3 \sin{\sqrt{2} \pi n}}$$

The residue at each pole $z=(\sqrt{2}+1) n \ne 0$ is

$$\frac{(-1)^n (\sqrt{2}+1)}{(\sqrt{2}+1)^3 \pi n^3 \sin{(\sqrt{2}+1) \pi n}} = \frac{(\sqrt{2}-1)^2}{\pi n^3 \sin{\sqrt{2} \pi n}}$$

Thus,

$$2 \sum_{n=1}^{\infty} \frac{1+(\sqrt{2}-1)^2}{\pi n^3 \sin{\sqrt{2} \pi n}} + \frac{13 \pi^2}{90} \left ( \sqrt{2}-1 \right ) = 0$$

because the contour integral is zero. Thus, I get that

$$\sum_{n=1}^{\infty} \frac{1}{n^3 \sin{\sqrt{2} \pi n}} = -\frac{13 \pi^3}{360 \sqrt{2}}$$

As a bonus, it looks like we have discovered a bug in Mathematica. As a matter of routine, I checked the result against a straight evaluation in Mathematica. To my horror, Mathematica returned $-13 \pi^{\color{red}{2}}/(360 \sqrt{2})$. How was I off by a factor of $\pi$? I checked and checked my work but found nothing wrong.

The solution to this problem lay in simply evaluating the sum numerically for an increasingly large number of terms. However, in order to assess whether there would be any surprises waiting for us from the sine term, I had to estimate the worst possible “spike” near an integer times $\pi$. What I found above is that, worst case, the terms decrease as some constant times $1/n^2$, so the effect of any spike is limited.

Armed with this information, I was able to verify in Mathematica that, indeed, numerical evaluations of finite sums converged to the answer I gave above rather than Mathematica’s result. Mr. Wolfram has received yet another letter.

20 Nov 2016

I just got the following email:

Hello Ron Gordon,

In Febuary 2016 you reported an issue with Mathematica wherein Sum returns a wrong answer for some expressions.
We believe that the issue has been resolved in the current release of Mathematica.

Thank you for your report and we look forward to a continued, productive relationship with you.

Best regards,

• Would you care to elaborate on the worst cast scenario being when $2n^2$ is next to a perfect square?
• Sure. If $2 n^2$ were a perfect square, then $\sin{\sqrt{2} \pi n}$ would be zero. But we know that $2 n^2$ can never be a perfect square, as $\sqrt{2}$ is irrational. Thus, $\left | \sin{\sqrt{2} \pi n} \right |$ is a minimum when the difference between $2 n^2$ and a perfect square $m^2$ is one, i.e., $2 n^2 = m^2 \pm 1$ as I stated in the answer.