MHB How Do You Solve the Alternating Series in POTW #283?

[Euge]

Ackbach has asked me to step in for him for a while. Here is this week's POTW:

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Evaluate the sum of the alternating series $$\sum\limits_{n = 1}^\infty \frac{(-1)^{n-1}}{n^4}$$
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Remember to read the http://www.mathhelpboards.com/showthread.php?772-Problem-of-the-Week-%28POTW%29-Procedure-and-Guidelines to find out how to http://www.mathhelpboards.com/forms.php?do=form&fid=2!

 

[Euge]

Congratulations to Opalg for his correct solution. You can read his solution below.

Spoiler

$$\sum_{n=1}^\infty\frac{(-1)^{n-1}}{n^4} = \sum_{n=1}^\infty\frac1{n^4} - 2\sum_{n=1}^\infty\frac1{(2n)^4} = \Bigl(1 - \frac2{16}\Bigr)\sum_{n=1}^\infty\frac1{n^4} = \frac{7\pi^4}{720},$$ using the famous fact that $$ \sum_{n=1}^\infty\frac1{n^4} = \zeta(4) = \frac{\pi^4}{90}.$$

One way to prove the formula $$\sum_{n=1}^\infty\frac1{n^4} = \frac{\pi^4}{90}$$ is to find the Fourier series for the function $f(x) = x^2$ on the interval $[-\pi,\pi].$

For $n\ne0$, $$\hat{\ f}(n) = \frac1{2\pi}\int_{-\pi}^\pi x^2e^{inx}dx = \frac1{2n\pi i}\Bigl[x^2e^{inx}\Bigr]_{-\pi}^\pi - \frac1{n\pi i}\int_{-\pi}^\pi xe^{inx}dx = 0 + \frac1{n^2\pi}\Bigl[xe^{inx}\Bigr]_{-\pi}^\pi - \frac1{n^2\pi}\int_{-\pi}^\pi e^{inx}dx = \frac{2(-1)^n}{n^2}$$ (integrating by parts twice). For $n=0$, $$\hat{\ f}(0) = \frac1{2\pi}\int_{-\pi}^\pi x^2dx = \frac1{2\pi}\Bigl[\frac{x^3}3\Bigr]_{-\pi}^\pi = \frac{\pi^2}3.$$

Now apply Parseval's theorem, which says that $$\|f\|_2^2 = \sum_{n\in\Bbb{Z}}|\hat{\ f}(n)|^2$$, where $$\|f\|_2^2 = \frac1{2\pi}\int_{-\pi}^\pi|f(x)|^2dx = \frac1{2\pi}\int_{-\pi}^\pi x^4 dx = \frac{\pi^4}5.$$ That gives $$\frac{\pi^4}5 = \frac{\pi^4}9 + 2\sum_{n=1}^\infty\frac4{n^4}$$, from which $$\sum_{n=1}^\infty\frac1{n^4} = \frac{\pi^4}{90}.$$