I was trying a problem from Ch -1( Elliptic Functions problem no. 15) of book Modular functions and Dirichlet series in number theory whose Statement is this.
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I have no idea on how to solve this problem, please give some hints. It is after the introduction to Lambert series in the exercises.
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I just wrote a longer reference-answer to a question that got closed, as it was a dublicate, so I post it here. I hope it is helpful.
In the book Number Theory 3 by the authors Nobushige Kurokawa, Masato Kurihara and Takeshi Saito in the first chapter on Modular Forms there are derived the values of this sort.
For example $$\sum_{n=1}^\infty \frac{n^5}{e^{2\pi n}-1}=\frac{1}{504}$$.
It seems that your question is not taken from this book, so you might want to check this out.
The following exercise is given, which is similar to what you are asking:
For $k\geq 6$ satisfying $k\equiv 2\mod 4$, find
$$\sum_{n=1}^\infty \frac{n^{k-1}}{e^{2\pi n}-1}$$
The book gives the following hint/answer:
The answer is $\dfrac{B_k}{2k}$, where $B_k$ is the Bernoulli number. This can be obtained as follows. By letting $z=i$ in the transformation formula $E_k(-\frac1z)=z^kE_k(z)$ of the holomorphic Eisenstein series of weight $k$, we obtain $E_k(i)=i^kE_k(i)=-E_k(i)$, which implies $E_k(i)=0$.
On the other hand, we have $$E_k(i)=1-\frac{k}{B_k}\sum_{n=1}^\infty \sigma_{k-1}(n)e^{-2\pi n}=1-\frac{2k}{B_k}\sum_{n=1}^\infty \frac{n^{k-1}}{e^{2\pi n}-1}$$
Where $\sigma_k(n)=\sum_{d|n} d^k$.
For part (a) you need another function $H(x) $ defined by $$H(x) =\sum_{n=1}^{\infty}\frac{n^5x^n}{1+x^n}$$ and you should note that $$G(x) - H(x)=2G(x^2)$$ and $$H(x) =F(x) +32H(x^2)$$ For part (b) you need some idea about elliptic function theory.
Your function $G$ is related to Ramanujan's function $R(q) $ via $$R(q) =1-504G(q)$$ If $k$ is the elliptic modulus corresponding to nome $q$ and $K$ is the corresponding complete elliptic integral of first kind then \begin{align} R(q)&=\left(\frac{2K}{\pi}\right) ^6(1+k^2)(1-34k^2+k^4)\notag\\ R(q^2)&=\left(\frac{2K}{\pi}\right) ^6(1+k^2)(1-2k^2)\left(1-\frac{k^2}{2}\right)\notag\\ R(q^4)&=\left(\frac{2K}{\pi}\right) ^6\left(1-\frac{k^2}{2}\right)\left(1-k^2-\frac{k^4}{32}\right)\notag \end{align} If $q=e^{-\pi} $ then $k^2=1/2$ and $$R(q) =-\left(\frac{2K} {\pi} \right) ^6\cdot\frac{3}{2}\cdot\frac{63}{4},R(q^2)=0,64R(q^4)=\left(\frac{2K}{\pi}\right)^6\cdot\frac{3}{2}\cdot\frac{63}{4}$$ and therefore $$R(q) - 34R(q^2)+64R(q^4)=0$$ or $$31-504\{G(q)-34G(q^2)+64G(q^4)\}=0$$ It follows that $$F(q) =\frac{31}{504}$$