Converting a Series with Even and Odd Terms into a Single Summation

[bolzano]

This is the last part to a Fourier Series problem. After using Parseval's Identity the following series emerges: \frac{1}{1^{4}}+\frac{1}{3^{4}} +\frac{1}{5^{4}} +\frac{1}{7^{4}}+\cdots = \frac{\pi^{4}}{96}

We are then asked to show: \frac{1}{1^{4}}+\frac{1}{2^{4}} +\frac{1}{3^{4}} +\frac{1}{4^{4}}+\cdots = \frac{\pi^{4}}{90}Must one split the left hand side of the last identity into \sum\frac{1}{(2n)^{4}} + \sum\frac{1}{(2n-1)^{4}}, or is there an easier way? I'm guessing there is.

Thanks a lot.

 

[Curious3141]

This is the last part to a Fourier Series problem. After using Parseval's Identity the following series emerges: \frac{1}{1^{4}}+\frac{1}{3^{4}} +\frac{1}{5^{4}} +\frac{1}{7^{4}}+\cdots = \frac{\pi^{4}}{96}

We are then asked to show: \frac{1}{1^{4}}+\frac{1}{2^{4}} +\frac{1}{3^{4}} +\frac{1}{4^{4}}+\cdots = \frac{\pi^{4}}{90}Must one split the left hand side of the last identity into \sum\frac{1}{(2n)^{4}} + \sum\frac{1}{(2n-1)^{4}}, or is there an easier way? I'm guessing there is.

Thanks a lot.

Hint:

Call your original series (sum of reciprocals of powers of odds) S. You're supposed to determine $\zeta(4)$

$\sum\frac{1}{(2n-1)^{4}} = S$

$\sum\frac{1}{(2n)^{4}} = \frac{1}{16}\zeta(4)$

and the sum of those two is equal to $\zeta(4)$ again, right?

Just rearrange and solve for $\zeta(4)$.

 

[bolzano]

But are you assuming that $\sum\frac{1}{(2n)^{4}} = \frac{1}{16}\zeta(4)$? Oh wait I think I understood. I'll let you know later thanks a lot.

 

[Curious3141]

But are you assuming that $\sum\frac{1}{(2n)^{4}} = \frac{1}{16}\zeta(4)$? Oh wait I think I understood. I'll let you know later thanks a lot.

You're just bringing the constant term $\frac{1}{2^4}$ outside the summation sign.