In one of his letters to G. H. Hardy, Ramanujan gave the following sum $$\dfrac{1}{1^{7}\cosh\left(\dfrac{\pi\sqrt{3}}{2}\right)} - \dfrac{1}{3^{7}\cosh\left(\dfrac{3\pi\sqrt{3}}{2}\right)} + \dfrac{1}{5^{7}\cosh\left(\dfrac{5\pi\sqrt{3}}{2}\right)} - \cdots = \frac{\pi^{7}}{23040}\tag{1}$$ or using $\sum $ notation $$\sum_{n = 0}^{\infty} \dfrac{(-1)^{n}}{(2n + 1)^{7}\cosh\left(\dfrac{(2n + 1)\pi\sqrt{3}}{2}\right)} = \frac{\pi^{7}}{23040}$$ Since $\cosh y = (e^{y} + e^{-y})/2$ we can see that the sum is equal to $$S = 2\sum_{n = 0}^{\infty}\dfrac{(-1)^{n}\exp\left(-\dfrac{(2n + 1)\pi\sqrt{3}}{2}\right)}{(2n + 1)^{7}\left\{1 + \exp\left(-(2n + 1)\pi\sqrt{3}\right)\right\}}$$ Putting $$q = \exp\left(-\pi\sqrt{3}\right)$$ we get the sum as $$S = 2\sum_{n = 0}^{\infty}\frac{(-1)^{n}q^{n + 1/2}}{(2n + 1)^{7}(1 + q^{2n + 1})}$$ We can then use $$\dfrac{q^{n + 1/2}}{1 + q^{2n + 1}} = \frac{q^{n + 1/2} - q^{3(n + 1/2)}}{1 - q^{4n + 2}}$$ and hope to use the Ramanujan functions $P, Q, R$ given by $$P(q) = 1 - 24\sum_{n = 1}^{\infty}\frac{nq^{2n}}{1 - q^{2n}}\\ Q(q) = 1 + 240\sum_{n = 1}^{\infty}\frac{n^{3}q^{2n}}{1 - q^{2n}}\\ R(q) = 1 - 504\sum_{n = 1}^{\infty}\frac{n^{5}q^{2n}}{1 - q^{2n}}$$ to calculate the sum $S$ in terms of $P, Q, R$. However the stumbling block is the term $(2n + 1)^{7}$ appearing in the denominator. Therefore I am not sure if the functions $P, Q, R$ can be used in the evaluation of sum $S$. Please let me know if there are any other approaches to calculate the sum $S$.
Update: While going through Ramanujan's Notebooks Vol 3 (by Bruce C. Berndt) I found the proof for the above sum. But he uses a highly non-obvious formula (also discovered by Ramanujan) $$\frac{u^{6n}}{\cos u\cos(\omega u)\cos(\omega^{2}u)} = 12\sum_{k = 0}^{\infty}\dfrac{(-1)^{k}\left\{\left(k + \dfrac{1}{2}\right)\pi\right\}^{6n + 5}}{\left[\left\{\left(k + \dfrac{1}{2}\right)\pi\right\}^{6} - u^{6}\right]\cosh\left\{\left(k + \dfrac{1}{2}\right)\pi\sqrt{3}\right\}}\tag{2}$$ where $\omega$ is a primitive cube root of unity. Berndt goes on to say that Ramanujan probably obtained this formula $(2)$ via partial fractions and says that it can be obtained by a routine procedure with somewhat lengthy calculation. The method of partial fractions is used to express a rational function (for the purpose of integrating them) in the form of a sum of finite terms which can be integrated via standard formulas. I am nor sure how that could be extended to any general function (which is not rational) and thereby produce a series. Berndt says "the sum $(1)$ can be obtained by putting $n = 0$ in $(2)$ and then equating coefficients of $u^{6}$ on both sides". This part requires some reasonable amount of calculation, but it is not so difficult.
I want to understand the technique of partial fractions as applied to general functions (may be with some requirement of continuity and differentiability) and its proper justification so that I can provide a proof of formula $(2)$ for myself and thereby have a complete proof of the Ramanujan's sum $(1)$.