How do I evaluate the sum $\sum_{n=1,3,5,...}^{\infty} \frac{1}{n^{4}}$?

Question

I'm trying to evaluate the following sum:

$\frac{480\hbar^{2}}{\pi^{4}ma^{2}} \cdot \sum_{n=1,3,5,...}^{\infty} \frac{1}{n^{4}}$

I've written out a couple partial sums for $n_{odd}$ up to $7$ : $1 + \frac{1}{81} + \frac{1}{625} + \frac{1}{2401}$ but am a bit lost as to where to go to actually evaluate this. I've looked online for the formula for the harmonic series and figured I would just try to parse the sum of the odd terms then raise that to the $4^{th}$ power... or do the same with the odd terms of $\frac{1}{n^{2}}$ and square them but I can't seem to find them or make any headway. Any help is much appreciated.

The solution is:

$\frac{5\hbar^{2}}{ma^{2}}$

Which tells me that the answer to my sum of odd numbered terms for $\frac{1}{n^{4}}$ must be $\frac{\pi^{4}}{96}$ but I still don't know how we got there. Any help is greatly appreciated!

Answer 1

Define Riemann zeta function as

$$ \zeta(s)=\sum_{n\ge1}{1\over n^s} $$

Then we have

\begin{aligned} \sum_{\substack{n\ge1\\n\text{ odd}}}{1\over n^s} &=\sum_{n\ge1}{1\over n^s}-\sum_{\substack{n\ge1\\n\text{ even}}}{1\over n^s} \\ &=\zeta(s)-\sum_{n\ge1}{1\over(2n)^s} \\ &=(1-2^{-s})\zeta(s) \end{aligned}

This indicates that

\begin{aligned} \frac{480\hbar^{2}}{\pi^{4}ma^{2}} \cdot \sum_{n=1,3,5,...}^{\infty} \frac{1}{n^{4}} &=\frac{480\hbar^{2}}{\pi^{4}ma^{2}}{15\over16}\zeta(4) \\ &={450\hbar^2\over\pi^4ma^2}\zeta(4) \end{aligned}

Finally, using the fact that $\zeta(4)={\pi^4\over90}$, we conclude

$$ \frac{480\hbar^{2}}{\pi^{4}ma^{2}} \cdot \sum_{n=1,3,5,...}^{\infty} \frac{1}{n^{4}}={5\hbar^2\over ma^2} $$

Answer 2

Another method is by using Fourier coefficients of the function $f$ on $[-\pi,\pi]$ defined by $$ f(x) = |x| $$ Then using integration by parts and using Fourier coefficient definition, one can obtain $$ \hat{f}(0) = \frac{\pi}{2} $$ and for $n \neq 0$, $$ \hat{f}(n) = \begin{cases} 0 & \text{ if $n$ is even} \\ \frac{-2}{n^2\pi} & \text{ if $n$ is odd} \\ \end{cases} $$ Now, the famous Parseval's identity (which relates function with it's Fourier coefficients) states that if $f \in L^2([-\pi,\pi])$, then

$$ \sum_{n = -\infty}^{\infty}|\hat{f}(n)|^2 = \frac{1}{2\pi}\int_{-\pi}^{\pi}|f(x)|^2dx $$ For our case, we can write

$$ |\hat{f}(0)|^2 + \sum_{n \in \mathbb{Z}\setminus \{0\}}|\hat{f}(n)|^2 = \frac{1}{2\pi}\int_{-\pi}^{\pi}x^2dx $$ Thus, $$ \frac{\pi^2}{4} + \sum_{n = \pm 1, \pm 2,...}\frac{4}{n^4\pi^2} = \frac{1}{2\pi}\frac{2\pi^3}{3} = \frac{\pi^2}{3} $$ which implies $$ \frac{4}{\pi^2} \cdot 2 \sum_{n=1,3,...}\frac{1}{n^4} = \frac{\pi^2}{12} $$ and hence $$ \sum_{n=1,3,...}\frac{1}{n^4} = \frac{\pi^2}{12}\cdot \frac{\pi^2}{8} = \frac{\pi^4}{96} $$

And hopefully you can use it in your computation.