Conjectured relation between hyperbolic sums

Question

Does $$ 2\sum_{k=1}^{\infty} \frac{(-1)^{k-1} k}{e^{2\pi k}-1} +\sqrt{2} \sum_{k=1}^{\infty} \frac{(-1)^{k-1}k}{e^{2\pi k}+1} \stackrel{?}{=} \frac{1}{4\pi} - \frac{2-\sqrt{2}}{8} $$ The two appear to agree numerically. Can someone confirm or link a reference? Thanks

Answer 1

Let $q=e^{-2\pi}$ and then the left hand side of the equation in question can be written as $$2\sum_{n\geq 1}\frac{(-1)^{n-1}nq^n}{1-q^n}+\sqrt{2}\sum_{n\geq 1}\frac{(-1)^{n-1}nq^n}{1+q^n}=2a(q)+\sqrt{2}b(q)\text{ (say)} \tag{1}$$ The functions $a(q), b(q) $ can be expressed in terms of Ramanujan's function $$P(q) =1-24\sum_{n\geq 1}\frac{nq^n}{1-q^n}\tag{2}$$ We have $$\sum_{n\geq 1}\frac{nq^n}{1-q^n}=\frac{1-P(q)}{24}$$ and therefore $$\sum_{n\geq 1}\frac{(-1)^{n-1}nq^n}{1-q^n}=\frac{1-P(q)}{24}-4\cdot\frac{1-P(q^2)}{24}$$ so that $$a(q) =\frac{4P(q^2)-P(q)}{24}-\frac{1}{8}\tag{3}$$ To evaluate $b(q) $ let us observe that $$\frac{1}{1-q^n}-\frac{1}{1+q^n}=\frac{2q^n}{1-q^{2n}}$$ and hence $$\sum_{n\geq 1}\frac{nq^n}{1+q^n}=\sum_{n\geq 1}\frac{nq^n}{1-q^n}-2\sum_{n\geq 1}\frac{nq^{2n}}{1-q^{2n}}=\frac{1-P(q)}{24}-\frac{2(1-P(q^2)}{24}=\frac{2P(q^2)-P(q)-1}{24}$$ And then we get $$ \sum_{n\geq 1}\frac{(-1)^{n-1}nq^n}{1+q^n}=\frac{2P(q^2)-P(q)-1}{24}-4\cdot \frac{2P(q^4)-P(q^2)-1}{24}$$ ie $$b(q) =\frac{6P(q^2)-8P(q^4)-P(q)}{24}+\frac{1}{8}\tag{4}$$

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Evaluation of the sum in question thus needs the evaluation of the function $P(q)$. For this purpose we need a little theory of the link between $P(q) $ and elliptic integrals.

Let $k\in(0,1)$ be the elliptic modulus and $k'=\sqrt{1-k^2}$ be the complementary modulus and we define complete elliptic integral $$K(k) =\int_0^{\pi/2}\frac{dx}{\sqrt{1-k^2\sin^2x}}\tag{5}$$ If $k$ is available from context then we use $K, K'$ to denote integrals $K(k), K(k')$. Corresponding to modulus $k$ we have nome $q=\exp(-\pi K'/K) $ and many functions of $q$ can be expressed in terms of $K, k$. Another key ingredient we need here is the Landen transformation which says that if $k$ is replaced by $(1-k')/(1+k')$ then $K$ changes to $(1+k')K/2$ and the ratio $K'/K$ gets doubled. Thus in other words if $q$ is the nome corresponding to $k$ then $q^2$ is the nome corresponding to $(1-k')/(1+k')$.

If we set $k=k'$ so that $k=1/\sqrt{2}$ then we have $K=K'$ and hence $q=e^{-\pi} $. Applying Landen transformation we see that $e^{-2\pi}$ is the nome corresponding to modulus $$\frac{1-k'}{1+k'}=3-2\sqrt{2}$$ In what follows we therefore use the values $$q=e^{-2\pi},k=3-2\sqrt{2}\tag{6}$$ The value of $P(q) =3/\pi$ is well known. We next need the formula $$2P(q^2)-P(q)=\left(\frac{2K}{\pi}\right)^2(1+k^2)\tag{7}$$ using which we get $$P(q^2)=\frac{3}{2\pi}+\left(\frac{2K}{\pi}\right)^2 (9-6\sqrt{2})$$ Applying Landen transformation on $(7)$ we get $$2P(q^4)-P(q^2)= \left(\frac{2K}{\pi}\right)^2\cdot\frac{1+k'^2}{2}$$ Adding twice of above equation to $(7)$ we get $$4P(q^4)-P(q)= 3\left(\frac{2K}{\pi}\right)^2\tag{8}$$ We then have $$P(q^4)=\frac{3}{4\pi}+\frac{3}{4} \left(\frac{2K}{\pi}\right)^2$$ Using the values of $P(q), P(q^2),P(q^4)$ we get the values of $a(q), b(q) $ from $(3),(4)$ and thereby the sum in question gets evaluated and matches the value in question. The nice thing to note here is that the expression $(2K/\pi)^2$ gets cancelled in the calculation and one does not need its explicit closed form evaluation in terms of gamma function.