A Ramanujan-type identity: $11\sum_{n=1}^{\infty}\frac{n^3}{e^{2n\pi}-1}-16\sum_{n=1}^{\infty}\frac{n^3}{e^{4n\pi}-1}=\frac{1}{48}$

Out of curiosity, why it is these sums yield a rational answer?

$$11\sum_{n=1}^{\infty}\frac{n^3}{e^{2n\pi}-1}-16\sum_{n=1}^{\infty}\frac{n^3}{e^{4n\pi}-1}=\frac{1}{48}$$

I found this identity during the observing ramanujan identiy via a wolfram sum calculator.

Or it is also related to the type of $e^{\pi{\sqrt43}}=884736743.999$ it is almost an integer?

Also,if possible can anyone verify it is correctness?

$$\sum_{n\geq 1}\frac{n^3}{e^{2n\pi}-1} = \sum_{n\geq 1}\sum_{k\geq 1} n^3 e^{-2k\pi n}=\frac{1}{8}\sum_{k\geq 1}\frac{3+2\sinh^2(k\pi)}{\sinh^4(k\pi)}\tag{1}$$ $$\sum_{n\geq 1}\frac{n^3}{e^{4n\pi}-1} = \sum_{n\geq 1}\sum_{k\geq 1} n^3 e^{-4k\pi n}=\frac{1}{8}\sum_{k\geq 1}\frac{3+2\sinh^2(2k\pi)}{\sinh^4(2k\pi)}\tag{2}$$ Now we may prove that: $$\sum_{k\geq 1}\frac{1}{\sinh^2(k\pi)}=\frac{1}{6}-\frac{1}{2\pi}.\tag{3}$$ Let we define $f(z)$ as $\frac{z}{e^{2\pi z}-1}$. $f$ has a removable singularity at the origin: by removing it, we are free to assume $f(0)=\frac{1}{2\pi}$. $f(z)$ has a remarkable property: $f(z)+\frac{z}{2}$ is an even function, hence in computing $$\sum_{n\geq 0}f(n) = \frac{1}{2\pi}+\sum_{n\geq 1}\frac{n}{e^{2\pi n}-1}$$ through the Euler-Maclaurin summation formula or through the Abel-Plana formula the remainder term greatly simplifies, leading to: $$\sum_{n\geq 1}\frac{n}{e^{2\pi n}-1}=\frac{1}{24}-\frac{1}{8\pi} \tag{4}$$ that proves $(3)$, since: $$\sum_{n\geq 1}\frac{n}{e^{2\pi n}-1} = \sum_{n\geq 1}\sum_{k\geq 1}n e^{-2k\pi n}=\frac{1}{4}\sum_{k\geq 1}\frac{1}{\sinh^2(k\pi)}.$$ In a similar fashion we get: $$\sum_{n\geq 1}\frac{1}{\sinh^4(n\pi)}=-\frac{11}{90}+\frac{1}{3 \pi }+\frac{\Gamma\left(\frac{1}{4}\right)^8}{1920 \pi ^6}\tag{5}$$ then by $(3)$ and $(5)$ it follows that:

$$\sum_{n\geq 1}\frac{n^3}{e^{2n\pi}-1} = \color{red}{\frac{\Gamma\left(\frac{1}{4}\right)^8}{5120 \pi ^6}-\frac{1}{240}} \tag{6}$$

and if we tackle $(2)$ in the same way, to prove the OP's claim is just a matter of trivial algebra.

Many thanks to Zucker's The Summation of Series of Hyperbolic Functions: the last term of $(5)$ comes from the computation of an elliptic integral - these objects are deeply related with series involving $\frac{1}{\sinh},\frac{1}{\cosh}$ or their powers.

• (+1) Very, very, very nice. I think someone should check if you are doped :-D well done bro. – Marco Cantarini May 15 '16 at 10:47

I provide an approach which involves many identities from the classical theory of elliptic and theta functions. Also I present them using Ramanujan's notation. Although the theory of elliptic and theta functions needs complex analysis to understand all the details, the formulas which I give here can be established within limits of real analysis only (and the proofs are available in my blog posts, let me know if you need them).

In what follows $k$ is a real number with $0 < k < 1$ and $k' = \sqrt{1 - k^{2}}$. Further $$K = K(k) = \int_{0}^{\pi/2}\frac{dx}{\sqrt{1 - k^{2}\sin^{2}x}}, K' = K(k'), E = E(k) = \int_{0}^{\pi/2}\sqrt{1 - k^{2}\sin^{2}x}\,dx$$ and $$q = \exp\left(-\pi\cdot\frac{K'}{K}\right)$$

Let us use Ramanujan Function $$Q(q) = 1 + 240\sum_{n = 1}^{\infty}\frac{n^{3}q^{n}}{1 - q^{n}}\tag{1}$$ Then we know that $$Q(q^{2}) = \left(\frac{2K}{\pi}\right)^{4}(1 - k^{2} + k^{4})\tag{2}$$ where $q = e^{-\pi K'/K}$ and we can change $q$ to $q^{2}$ if $k$ is replaced by $(1 - k')/(1 + k')$ and $K$ is replaced by $K(1 + k')/2$ Thus we have $$Q(q^{4}) = \left(\frac{K(1 + k')}{\pi}\right)^{4}\left(1 - \frac{(1 - k')^{2}}{(1 + k')^{2}} + \frac{(1 - k')^{4}}{(1 + k')^{4}}\right)$$ or $$Q(q^{4}) = \left(\frac{2K}{\pi}\right)^{4}\left(1 - k^{2} + \frac{k^{4}}{16}\right)\tag{3}$$ Lets put $q = e^{-\pi}$ to get $K' = K$ so that $k = 1/\sqrt{2}$. Then $$\sum_{n = 1}^{\infty}\frac{n^{3}}{e^{2n\pi} - 1} = \frac{Q(q^{2}) - 1}{240}, \sum_{n = 1}^{\infty}\frac{n^{3}}{e^{4n\pi} - 1} = \frac{Q(q^{4}) - 1}{240}$$ so the desired sum is $$\frac{1}{240}\{5 + 11Q(q^{2}) - 16Q(q^{4})\} = \frac{1}{48} + \frac{(2K/\pi)^{4}}{240}\{11\cdot(3/4) - 16\cdot(33/64)\} = \frac{1}{48}$$

About the values of such sums being rational I don't see them as surprising. Clearly for $q = e^{-\pi}$ we have $$Q(q^{2}) = \frac{3}{4}\left(\frac{2K}{\pi}\right)^{4}$$ And we can connect it with many other formulas like $$P(q) = 1 - 24\sum_{n = 1}^{\infty}\frac{nq^{n}}{1 - q^{n}} = \left(\frac{2K}{\pi}\right)^{2}\left(\frac{6E}{K} + k^{2} - 5\right)$$ and $$P(q^{2}) = \left(\frac{2K}{\pi}\right)^{2}\left(\frac{3E}{K} + k^{2} - 2\right)$$ so that $$2P(q^{2}) - P(q) = \left(\frac{2K}{\pi}\right)^{2}\left(1 + k^{2}\right) = \frac{3}{2}\left(\frac{2K}{\pi}\right)^{2}$$ and hence finally $$3Q(q^{2}) - \{2P(q^{2}) - P(q)\}^{2} = 0$$ And this will give a polynomial equation with rational coefficients connecting $P(q), P(q^{2}), Q(q^{2})$. Many such relations can be derived via simple manipulation of the formulas for $P, Q, R$ in terms of $k, K, E$. Just to add one last example of such sums with rational value, it can be established using some of Ramanujan's results (or some other techniques which I am not aware of) that $$\sum_{n = 1}^{\infty}\frac{n^{s}}{e^{n\pi} - 1}$$ is rational for all positive integers $s$ of the form $s = 4k + 1$ (the sum has a closed form for odd values of $s$ but not for even values of $s$).

In his paper "Modular Equations and Approximations to $\pi$" Ramanujan does the ultimate form of cancellation to obtain various series for $1/\pi$ some of which consists purely of rational terms and the most spectacular one consists of just one irrational $\sqrt{2}$ namely $$\boxed\frac{1}{2\pi\sqrt{2}} = \frac{1103}{99^{2}} + \frac{27493}{99^{6}}\frac{1}{2}\frac{1\cdot 3}{4^{2}} + \frac{53883}{99^{10}}\frac{1\cdot 3}{2\cdot 4}\frac{1\cdot 3\cdot 5\cdot 7}{4^{2}\cdot 8^{2}} + \cdots$$ See the series on my blog starting with this post for more details.

• And things just happen to cancel out. Nice. – marty cohen May 14 '16 at 15:49
• @martycohen: The coefficients $11$ and $16$ were chosen to cancel things nicely. – Paramanand Singh May 14 '16 at 16:23
• This is a truely epic answer, it is a pity I can give just one (+1). Slick way to avoid $\Gamma\left(\frac{1}{4}\right)^8$. – Jack D'Aurizio May 14 '16 at 16:46
• thanks a lot @JackD'Aurizio: even a $\delta$ of positive feedback from you is too much to handle and you are using "epic". you made my day just by giving this comment!!! – Paramanand Singh May 14 '16 at 16:53
• @JackD'Aurizio: BTW the factor of $\Gamma(1/4)$ is related to the value of $K(1/\sqrt{2})$ namely $K(1/\sqrt{2}) = (1/4\sqrt{\pi})\Gamma^{2}(1/4)$ – Paramanand Singh May 14 '16 at 16:58