# infinite series $n^7/(\exp(2\pi n)-1)$

I found an interesting topic on this site with regards to the series I am trying to evaluate:

Summing $\frac{1}{e^{2\pi}-1} + \frac{2}{e^{4\pi}-1} + \frac{3}{e^{6\pi}-1} + \cdots \text{ad inf}$

I was wondering if there is a closed form for even m when we have:

$$\sum_{n=1}^{\infty}\frac{n^{2m-1}}{e^{2\pi n}-1}$$

I have the series $$\sum_{n=1}^{\infty}\frac{n^{7}}{e^{2\pi n}-1},$$ I am trying to evaluate.

The aforementioned thread mentions that if m > 1 and odd, then we can use

$$\frac{B_{2m}}{4m}$$ to find the sum.

But, if m is even, the formula omits and error term.

Does anyone have info on this error term or how to evaluate my series or others that involve even m?.

I noticed that when m=1, there is an error term of $$\frac{-1}{8\pi}$$

The error appears to get smaller the larger m, and thus the power of n, becomes.

Thanks to all.

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I see you have a 0% acceptance rate... If you post a question and receive a satisfactory answer, you can (and should) "accept" one of the satisfactory answers by clicking the check mark which can be found hear the up/down votes for each answer. –  Tyler Mar 28 '11 at 4:50

Take a look at equation $(6)$ in Ramanujan's Formula for the Logarithmic Derivative of the Gamma Function by David Bradley.

Note that just before the formula it says, "Let $N$ be a positive integer," but the formula is valid for negative $N$ as well.

When $N$ is a negative even integer, where $N=-2m,$ the sum on the RHS of $(6)$ is taken to be the empty sum, and so is equal to zero. This means that for $N=-4$ we have
$$\sum_{k=1}^\infty \frac{k^7}{ e^{2\pi k} -1 } = \frac{\pi}{8} \sum_{k=1}^\infty \frac{k^8}{ \sinh^2(\pi k) } - \frac{1}{480}$$
$$\sum_{k=1}^\infty \frac{k^{4m-1}}{ e^{2\pi k} -1 } = \frac{\pi}{4m} \sum_{k=1}^\infty \frac{k^{4m}}{ \sinh^2(\pi k) } + \frac{B_{4m}}{8m}.$$
Thanks for the response, Derek. That sure is a long and messy looking formula. Ramanujan was something else. Anyway, I thought I would isolate the series at hand from that formula. The thing is, I need N=-4 in the formula. How would one use a negative term in a summation. The right side of [6] would be $$2\pi\sum_{k=0}^{-3}(-1)^{k+1}(2k-1)\frac{B_{2k}}{(2k)!}\frac{B_{-2-2k}}{(-2-2k‌​)!}$$. Same thing for the Bernoulli numbers. I see it says it works for negative N, but I do not see how. See what I am getting at?. Thanks again. –  Cody Jan 29 '11 at 0:43