Question on an infinite summation series

[phymath7]

Is the infinite series $\sum_{n=1,3,5,...}^\infty \frac {1} {n^6}$ somewhat related to the Riemann zeta function?The attached image suggest the value to be inverse of the co-efficient of the series.Is there any integral representation of the series from where the series can be evaluated?

 

[PeroK]

I think these identities can be calculated using an appropriate Fourier series expansion:

https://math.stackexchange.com/ques...y-frac1n6-frac-pi6945-by-fourier-series-of-x2

 

[pasmith]

These series are usually evaluated by applying Parseval's Theorem to in this case $x^3$ on $[-\pi,\pi]$. But you should perhaps look at reference 20, and also see how the $c_n$ were defined in the first place.

(If you have questions about an extract from another book or document, please tell us where it came from so we can consult the original in its context, if we have it available.)

When the sum converges absolutely, we have $$\sum_{\mbox{odd n}} \frac1{n^z} = \sum_{n=1}^\infty \frac1{n^z} - \sum_{\mbox{even n}} \frac1{n^z}.$$ Since $$\sum_{\mbox{even n}} \frac1{n^z} = \sum_{r=1}^\infty \frac1{(2r)^z} = 2^{-z} \sum_{r=1}^\infty \frac1{r^z}$$ we have $$\sum_{\mbox{odd n}} \frac1{n^z} = (1 - 2^{-z}) \sum_{n=1}^\infty \frac1{n^z}.$$

 

[julian]

Is the infinite series $\sum_{n=1,3,5,...}^\infty \frac {1} {n^6}$ somewhat related to the Riemann zeta function?The attached image suggest the value to be inverse of the co-efficient of the series.Is there any integral representation of the series from where the series can be evaluated?

It is closely connected to the Riemann zeta function, as noted by @pasmith. Such summations, of the form $\sum_{\text{odd } n} \frac{1}{n^{2k}}$, can be evaluated directly without needing Fourier series knowledge, using a complex contour integration technique. I have attached some notes detailing this method, where I've derived a number of established results (draft version), including the closed formula for $\zeta(2k)$ in terms of Bernoulli numbers:

\begin{align*}
\zeta(2k) = \sum_{n=1}^\infty \frac{1}{n^{2k}} = (-1)^{k+1} (2 \pi)^{2k} \dfrac{B_{2k}}{2 (2k)!}
\end{align*}