Entry 23

For powers \(k=4n+3\), Ramanujan gave $$\sum_{n=1} \frac{n^{3}}{e^{2n\pi}-1} = \frac{1^{3}}{e^{2\pi}-1}+\frac{2^{3}}{e^{4\pi}-1}+\frac{3^{3}}{e^{6\pi}-1}+\cdots=\frac{\Gamma\big(\tfrac{1}{4}\big)^8}{2^{10}\cdot5\pi^6}-\frac{1}{240}$$ $$\sum_{n=1} \frac{n^{7}}{e^{2n\pi}-1} = \frac{1^{7}}{e^{2\pi}-1}+\frac{2^{7}}{e^{4\pi}-1}+\frac{3^{7}}{e^{6\pi}-1}+\cdots=\frac{3\,\Gamma\big(\tfrac{1}{4}\big)^{16}}{2^{17}\cdot5\pi^{12}}-\frac{1}{480}$$ $$\sum_{n=1}^\infty  \frac{n^{11}}{e^{2n\pi}-1} = \frac{1^{11}}{e^{2\pi}-1}+\frac{2^{11}}{e^{4\pi}-1}+\frac{3^{11}}{e^{6\pi}-1}+\cdots =\frac{189\,\Gamma\big(\tfrac{1}{4}\big)^{24}}{2^{22}\cdot5\cdot13\,\pi^{18}}-\frac{691}{65520}$$ while for \(k=4n+1\) it evaluates to a rational number $$\sum_{n=1} \frac{n^{5}}{e^{2n\pi}-1} = \frac{1^{5}}{e^{2\pi}-1}+\frac{2^{5}}{e^{4\pi}-1}+\frac{3^{5}}{e^{6\pi}-1}+\cdots=\frac{1}{504}$$ $$\sum_{n=1} \frac{n^{9}}{e^{2n\pi}-1}=\frac{1^{9}}{e^{2\pi}-1}+\frac{2^{9}}{e^{4\pi}-1}+\frac{3^{9}}{e^{6\pi}-1}+\cdots=\frac{1}{264}$$ $$\sum_{n=1} \frac{n^{13}}{e^{2n\pi}-1}=\frac{1^{13}}{e^{2\pi}-1}+\frac{2^{13}}{e^{4\pi}-1}+\frac{3^{13}}{e^{6\pi}-1}+\cdots=\frac{1}{24}$$ and so on. The case \(k=1\) diverges.