TITLE: Show that $\sum^\infty_{k=0} \frac{2(-1)^k}{(2k+1)\pi\cosh[(2k+1)\pi/2]}=1/4$
QUESTION [17 upvotes]: Title says it all.
Well, maybe some backstory.

Flipping through my past notebooks, I found this:

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

[end of page]

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?

REPLY [1 votes]: Let us consider the function $$F(q) =\sum_{n=0}^{\infty} \frac{(-1)^nq^{n+(1/2)}}{(2n+1)(1+q^{2n+1})}\tag{1}$$ and then the question asks us to show $F(e^{-\pi}) =\pi/16$.
Let $k$ be the elliptic modulus corresponding to nome $q$ so that $q=e^{-\pi K'/K} $ where $K, K'$ are complete elliptic integrals of first kind corresponding to modulus $k$ and complementary modulus $k'=\sqrt{1-k^2}$. And then we prove that $$F(q) =\frac{1}{4}\arcsin k\tag{2}$$ This formula is originally due to Jacobi and one can provide a proof based on Fourier series of elliptic functions. The current problem is an instance of this formula for $q=e^{-\pi}, k=1/\sqrt{2}$ giving us $F(e^{-\pi}) =\pi/16$.
Let us observe that $$F(q) =\sum_{n=0}^{\infty} (-1)^n\frac{q^{n+(1/2)}}{2n+1}\sum_{i=0}^{\infty} (-1)^iq^{(2n+1)i}=\sum_{i=0}^{\infty} (-1)^i\sum_{n=0}^{\infty} (-1)^n\frac{q^{(i+(1/2))(2n+1)}}{2n+1}=\sum_{i=0}^{\infty} (-1)^i\arctan(q^{i+(1/2)})$$ Let us differentiate the above with respect to $k$ to get $$\frac{dq} {dk} \frac{dF(q)} {dq} =\frac{dq} {dk} \frac{1}{2}\sum_{n=0}^{\infty}(-1)^n \frac{(2n+1)q^{n-(1/2)}}{1+q^{2n+1}}$$ Next we note that $$\frac{dq} {dk} =\frac{\pi^2q}{2kk'^2K^2}$$ and hence above expression is transformed into $$\frac{dF} {dk} =\frac{\pi^2}{4kk'^2K^2}\sum_{n=0}^{\infty} (-1)^n\frac{(2n+1)q^{n+(1/2)}}{1+q^{2n+1}}\tag{3}$$ The sum above looks familiar and is related to Fourier series of elliptic functions. We have $$\operatorname{sd} (u, k) =\frac{2\pi}{kk'K}\sum_{n=0}^{\infty} (-1)^n\frac{q^{n+(1/2)}}{1+q^{2n+1}}\sin((2n+1)x)\tag{4}$$ where $u=(2K/\pi) x$ (see page 511, A Course of Modern Analysis, Whittaker and Watson).
Differentiating the above with respect to $x$ we get $$\frac{2K}{\pi}\frac{\operatorname {cn} (u, k)} {\operatorname {dn} ^2(u,k)}=\frac{2\pi}{kk'K}\sum_{n=0}^{\infty} (-1)^n\frac{(2n+1)q^{n+(1/2)}}{1+q^{2n+1}}\cos((2n+1)x)$$ Putting $u=0$ (so that $x=0$) we get $$\sum_{n=0}^{\infty} (-1)^n\frac{(2n+1)q^{n+(1/2)}}{1+q^{2n+1}}=\frac{kk'K^2}{\pi^2}\tag{5}$$ And then from $(3)$ and $(5)$ we get $$\frac{dF} {dk} =\frac{1}{4k'}=\frac{1}{4\sqrt{1-k^2}}$$ Integrating with respect to $k$ and noting that $F(q) =0$ when $k=0$ we get $$F(q) =\frac{1}{4}\arcsin k$$ Using different values of $q$ we can also obtain, among many examples, $$F(e^{-\pi\sqrt{3}}) = \frac{\pi}{48} \qquad F(e^{-\pi/\sqrt{3}}) = \frac{5\pi}{48}$$
$$F(e^{-\pi\sqrt{2}}) = \frac{1}{4}\arcsin(\sqrt{2}-1) \qquad F(e^{-2\pi}) = \frac{1}{4}\arcsin(3-2\sqrt{2})$$

Note: A similar question was asked today, 5 years later.
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