# Show that $\sum^\infty_{k=0} \frac{2(-1)^k}{(2k+1)\pi\cosh[(2k+1)\pi/2]}=1/4$

Title says it all.

Well, maybe some backstory.

Flipping through my past notebooks, I found this:

$$\vdots$$ $$= \sum^\infty_{k=0} \frac{2(-1)^k}{(2k+1)\pi\cosh[(2k+1)\pi/2]}\\=\frac14\quad(?)$$

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Ah, yes. My engineering numerical analysis professor gave me one homework problem:

Make a contour plot of the steady-state temperature distribution of a square plate, with the temperature of one of its side maintained at $1$ unit, the other three sides maintained at $0$ units.

Steady-state temperature distribution $T$ follows $\nabla^2T=0$.

I was calculating the temperature of the center of the plate analytically to sanity-check my program output, when I encountered this weird-looking sum. I remember I made a spreadsheet to numerically evaluate the sum, and Excel said 0.2500000... So I jot down $\frac14$ with a question mark beside it. The next page is filled with my fruitless attempts to find out the exact sum, to convince myself that the sum is exactly $1/4$.

I posted a bounty on my social media accounts, whoever can show the proof (or disproof) gets free beer.

After a few days, the bounty was still unclaimed, but I realized, "duuuuuhhhh! I could just exploit the symmetry. If I rotate the plate $90^\circ$ three times and superpose all the temperature distribution with the original one, temperature is $1$ unit everywhere. By linearity and symmetry, the temperature of the center point is therefore $1/4$."

I treated myself a few rounds of beer, and moved on. Case closed.

Looking again at this, I now realize that I might be missing out on some summation tactics I never knew.

Anybody can show me how is this equal to $1/4$ just by algebraic manipulation?

• +1 for the story, +1 more for the solution, and +1 even more for trying to learn greater things. – Simply Beautiful Art Jan 12 '17 at 23:22
• The residue theorem might come in useful. – Michael M Jan 13 '17 at 0:23
• Thanks for the suggestion and the upvote. Probably will take me a while to work that out, as I never took a course in analysis. But now I realize that math is wonderful, there are always more beautiful things to learn. – f1garo Jan 13 '17 at 1:25
• – Math-fun Jan 13 '17 at 9:49

I am busy as hell today but this is such a nice problem (and amusingly posed question) i can't resist... :)

The key idea here is simple: Look for a function in the complex plane $f(z)$ which has poles at the correct values $z_n$ and integrate it over an appropriate contour $C$.

It turns out that the function (which is also the simplest guess i can imagine)

$$f(z)=\frac{1}{z \cos(z)\cosh(z)}$$

is the correct choice. Its poles are given by $z_0=0$, $z_k=\frac{(2 k-1)\pi}{2}$ and $\tilde{z}_k=i \frac{(2 k-1)\pi}{2}$ with $k \in \mathbb{Z/0}$. We furthermore have

$$\text{res}(z_0)=1 \\ \text{res}(z_k)=\text{res}(\tilde{z}_k)=\frac{2}{\pi}\frac{(-1)^{k}}{(2k-1)\cosh\left(\pi\frac{2k-1}{2}\right)}\\ \text{res}(z_{-k})=\text{res}(z_k)$$

since $z\cos(z)\cosh(z)\sim |z| e^{a |z|}$ with $\Re(a)>0$ as $|z|\rightarrow\infty$ we can choose a big circle traversed anticlockwise with radius choosen such that we hit no pole of $f(z)$ as the integration contour $C$ (Thanks to @Dr. MV for rigorizing this point).

Then, we get by applying the residue theorem, we get in the limit of infinte radius

$$\oint_Cf(z)dz=2\pi i\text{res}(z_0)+4\pi i \sum_{k\geq1}\text{res}(z_k)+4\pi i\sum_{k\geq1}\text{res}(\tilde{z}_k)=0$$

or

$$8\pi i \sum_{k\geq1}\text{res}(z_k)=-2\pi i$$

which can be rewritten as (after shifting $k\rightarrow k+1$)

$$\sum_{k\geq0}\frac{2}{\pi}\frac{(-1)^{k}}{(2k+1)\cosh\left(\pi\frac{2k+1}{2}\right)}=\frac{1}{4}\\ \textbf{Q.E.D.}$$

To make things clearer, here is a sketch of the integration contour $\color{blue}{C}$ and the singularities $\color{red}{z_n}$

• (+1) This is extremely close to Ramanujan original argument, I'd say it is perfectly fine! – Jack D'Aurizio Jan 13 '17 at 12:51
• @JackD'Aurizio thanks! This is a relief...still wondering how this complicated sum can be derived with nearly zero effort – tired Jan 13 '17 at 12:53
• A big (+1). I was working on this last night, had identified $f(z)$ and then floundered to find a suitable contour. Now, I feel stupid. Well done my friend! -Mark – Mark Viola Jan 13 '17 at 14:42
• I was trying to isolate the poles on the real axis by taking a rectangular contour in hopes that the top and bottom halves would cancel. They didn't. I even recognized the symmetry between the residues on real and imaginary axes. For some reason (it was getting late and I was tiring) I just couldn't see the obvious "big circle" contour was the way forward. One note; we must take the limit with discrete radii to ensure the contour doesn't hit a pole. – Mark Viola Jan 13 '17 at 14:55
• In general, the function $\pi \sec(\pi z) f(z)$ can be used to evaluate infinite sums of the form $\sum_{n \in \mathbb{Z}} (-1)^n f(n+1/2)$, while the function $\pi \tan(\pi z) f(z)$ can be used to evaluate infinite sums of the form $\sum_{n \in \mathbb{Z}} f(n+1/2)$. In some cases, of course, the contour integral won't vanish in the limit. – Random Variable Jan 13 '17 at 19:56

As mentioned by Zucker in The summation of series of hyperbolic functions, in Question 358, J. Indian Math. Soc., 4 (1912), p. 78. Ramanujan proves (through the residue theorem) that $$\sum_{n\geq 1}(-1)^{n+1}(2n-1)^{4m-1}\,\text{sech}\left[(2n-1)\frac{\pi}{2}\right] = 0$$ holds for every $m\geq 0$. Your identity then follows from the $m=0$ case.

It is truly remarkable to know that such identity can be proved through Dirichlet's problem about the eigenvalues of the Laplacian operator. Can you be more specific about the relation between such a series and the physical problem?$^{(0)}$ A brilliant piece of math might arise from there.

$^{(0)}$Update: I found the connection - such series are related with the Green function of a square.

I will show an interesting technique for deriving a similar identity.

From the Weierstrass product $$\cosh(\pi x/2) = \prod_{m\geq 0}\left(1+\frac{z^2}{(2m+1)^2}\right)\tag{1}$$ by applying $\frac{d^2}{dz^2}\log(\cdot)$ to both sides we get: $$\frac{\pi^2}{8\cosh^2(\pi x/2)}=\sum_{m\geq 0}\frac{(2m+1)^2-z^2}{((2m+1)^2+z^2)}\tag{2}$$ If we replace $z$ with $(2n+1)$ and sum over $n\geq 0$, $$\sum_{n\geq 0}\frac{\pi^2}{8\cosh^2(\pi(2n+1)/2)} = \sum_{n\geq 0}\sum_{m\geq 0}\frac{(2m+1)^2-(2n+1)^2}{((2m+1)^2+(2n+1)^2)^2} \tag{3}$$ where the RHS of $(3)$ can also be written as $$\sum_{n\geq 0}\sum_{m\geq 0}\int_{0}^{+\infty}\cos((2n+1)x)x e^{-(2m+1)x}\,dx = \sum_{n\geq 0}\int_{0}^{+\infty}\frac{x\cos((2n+1)x)}{2\sinh(x)}\,dx\tag{4}$$ or, by exploiting integration by parts, $$\sum_{n\geq 0}\int_{0}^{+\infty}\frac{x\cosh(x)-\sinh(x)}{2\sinh(x)}\cdot\frac{\sin((2n+1)x)}{2n+1}\,dx\tag{5}$$ On the other hand, $\sum_{n\geq 0}\frac{\sin((2n+1)x)}{2n+1}$ is the Fourier series of a $2\pi$-periodic rectangle wave that equals $\frac{\pi}{4}$ over $(0,\pi)$ and $-\frac{\pi}{4}$ over $(\pi,2\pi)$. That implies, by massive cancellation: $$\sum_{n\geq 0}\frac{1}{\cosh^2(\pi(2n+1)/2)}=\frac{1}{2\pi}.\tag{6}$$ On the other hand, the Fourier transform of $\frac{1}{\cosh^2(\pi x)}$ is given by by $\frac{s\sqrt{8\pi}}{\sinh(\pi s)}$.
By Poisson's summation formula, $$\sum_{n\geq 1}\frac{(-1)^{n+1}n}{\sinh(\pi n)}=\frac{1}{4\pi}.\tag{7}$$

• Hey Jack, since i can't read the source you mentioned, could you please check if their derivation fits my approach? i'm quiet unsure if i made no big mistake here (it seems to easy to be true). thanks! – tired Jan 13 '17 at 12:38
• For the pyhsical background you might be interested in this presentation: cedricthieulot.net/diva/05.pdf. The symmetry considerations mentioned by op are essentially saying that: By superposition, adding four plates with one side on a different temperature then the other three (and center temperature given by the complicated sum $S$) gives one plate with constant temperature with core temerature is trivially one. From this $S=1/4$ follows directly – tired Jan 13 '17 at 13:03
• It looks like my predictions were true: sciencedirect.com/science/article/pii/S0955799706000683 – Jack D'Aurizio Jan 13 '17 at 13:12
• @tired: thank you. I always find very pleasing to work on some problem with you. – Jack D'Aurizio Jan 13 '17 at 13:13
• D'Aurizo Yeah, this is a nice proof of concept that physicists and mathematicans CAN work together ;-) – tired Jan 13 '17 at 13:17