1. ## Show the sum

Hi, can anyone help me with this question please?

Show that:
$\sum_{n=0}^{\infty} \frac{1}{(2n+1)^6}=\frac{\pi^6}{960}$

2. I just got a bit idea.
Should I try to do something about partial sum? Then do something about showing the series is converging?
But I don't know how to do it.

3. Originally Posted by tsang
Hi, can anyone help me with this question please?

Show that:
$\sum_{n=0}^{\infty} \frac{1}{(2n+1)^6}=\frac{\pi^6}{960}$

$\displaystyle 1 + \frac{1}{3^6} + \frac{1}{5^6} + .... = \left( 1 + \frac{1}{2^6} + \frac{1}{3^6} + \frac{1}{4^6} + .... \right) - \left(\frac{1}{2^6} + \frac{1}{4^6} + .... \right) = \left( 1 + \frac{1}{2^6} + \frac{1}{3^6} + \frac{1}{4^6} + .... \right) - \frac{1}{2^6} \left(1 + \frac{1}{2^6} + .... \right)$

That is,

$\displaystyle S = \left( 1 + \frac{1}{2^6} + \frac{1}{3^6} + \frac{1}{4^6} + .... \right) - \frac{S}{2^6}$

and $\displaystyle 1 + \frac{1}{2^6} + \frac{1}{3^6} + \frac{1}{4^6} + ....$ is a well known series .... (Google it).

A more rigorous proof can be constructed using Fourier analysis.

4. Originally Posted by mr fantastic
$\displaystyle 1 + \frac{1}{3^6} + \frac{1}{5^6} + .... = \left( 1 + \frac{1}{2^6} + \frac{1}{3^6} + \frac{1}{4^6} + .... \right) - \left(\frac{1}{2^6} + \frac{1}{4^6} + .... \right) = \left( 1 + \frac{1}{2^6} + \frac{1}{3^6} + \frac{1}{4^6} + .... \right) - \frac{1}{2^6} \left(1 + \frac{1}{2^6} + .... \right)$

That is,

$\displaystyle S = \left( 1 + \frac{1}{2^6} + \frac{1}{3^6} + \frac{1}{4^6} + .... \right) - \frac{S}{2^6}$

and $\displaystyle 1 + \frac{1}{2^6} + \frac{1}{3^6} + \frac{1}{4^6} + ....$ is a well known series .... (Google it).

A more rigorous proof can be constructed using Fourier analysis.

Thank you so much.
May I ask how to use Fourier analysis please? Could you please give me some details? I'm quite bad with Fourier series, it has been one weakness for me for a long time.
Thanks a lot.

5. Originally Posted by mr fantastic
$\displaystyle 1 + \frac{1}{3^6} + \frac{1}{5^6} + .... = \left( 1 + \frac{1}{2^6} + \frac{1}{3^6} + \frac{1}{4^6} + .... \right) - \left(\frac{1}{2^6} + \frac{1}{4^6} + .... \right) = \left( 1 + \frac{1}{2^6} + \frac{1}{3^6} + \frac{1}{4^6} + .... \right) - \frac{1}{2^6} \left(1 + \frac{1}{2^6} + .... \right)$

That is,

$\displaystyle S = \left( 1 + \frac{1}{2^6} + \frac{1}{3^6} + \frac{1}{4^6} + .... \right) - \frac{S}{2^6}$

and $\displaystyle 1 + \frac{1}{2^6} + \frac{1}{3^6} + \frac{1}{4^6} + ....$ is a well known series .... (Google it).

A more rigorous proof can be constructed using Fourier analysis.

How can I prove in most easy way that $\Sigma^\infty _{n=1} \frac{1}{n^6}=\frac{\pi^6}{945}$?

6. Originally Posted by tsang
Thank you so much.
May I ask how to use Fourier analysis please? Could you please give me some details? I'm quite bad with Fourier series, it has been one weakness for me for a long time.
Thanks a lot.
Use the Fourier series for the half-range even periodic extension of the function $f(t) = -t^2 + 2t$ where $0 \leq t \leq 1$ and apply Parseval's Theorem. (And no, I'm not going to do the calculation, that's your job if you want it).