By using this answer (On proving that $\sum\limits_{n=1}^\infty \frac{n^{13}}{e^{2\pi n}-1}=\frac 1{24}$), I found that $$2\frac{1-2^{-4k-2}}{1-2^{-4k-1}}\frac{(4k+2)!}{(2\pi)^{4k+2}(8k+4)}\lt \sum_{n=1}^\infty \frac{n^{4k+1}}{e^{2\pi n}-1}\lt \frac{2}{1-2^{-4k-1}}\frac{(4k+2)!}{(2\pi)^{4k+2}(8k+4)}$$ where $k\in\mathbb{N}$.
Are there some known bounds (useful for approximations) for the similar series $$\sum_{n=1}^\infty \frac{n^{4k-1}}{e^{2\pi n}-1}$$ where $k\in\mathbb{N}$?
