# Problem Of The Week # 317 - Aug 01, 2018

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#### Euge

##### MHB Global Moderator
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Congratulations to Opalg for his correct solution. Honorable mention goes to greg1313 and kaliprasad for proofs that used the value of $\zeta(4)$ without proof. You can read Opalg 's solution below.

Let $f(x) = x^2$, on the interval $[-\pi,\pi]$. The Fourier coefficients of $f$ are given by $$\displaystyle \hat{\ f}(n) = \frac1{2\pi}\int_{-\pi}^\pi x^2e^{-inx}dx.$$

When $n=0$, that gives $$\displaystyle \hat{\ f}(0) = \frac1{2\pi}\int_{-\pi}^\pi x^2dx = \frac1{2\pi}\left[ \frac{x^3}3\right]_{-\pi}^\pi = \frac{\pi^2}3.$$

For $n\ne0$, integrate by parts twice to get \begin{aligned} \hat{\ f}(n) &= \frac1{2\pi}\int_{-\pi}^\pi x^2e^{-inx}dx \\ &= \frac1{-2in\pi}\left[ x^2e^{-inx}\right]_{-\pi}^\pi - \frac i{n\pi}\int_{-\pi}^\pi xe^{-inx}dx \\ &= 0 + \frac1{n^2\pi}\left[ xe^{-inx}\right]_{-\pi}^\pi - \frac1{n^2\pi} \int_{-\pi}^\pi e^{-inx}dx \\ &= \frac{2(-1)^n}{n^2} .\end{aligned} By Parseval's theorem, $$\displaystyle \frac1{2\pi}\int_{-\pi}^\pi |f(x)|^2dx = \sum_{\Bbb N}|\hat{\ f}(n)|^2.$$ Since $$\displaystyle \int_{-\pi}^\pi |f(x)|^2dx = \int_{-\pi}^\pi x^4dx = \frac{2\pi^5}5,$$ it follows that $$\frac{\pi^4}5 = |\hat{\ f}(0)|^2 + 2\sum_{n=1}^\infty|\hat{\ f}(n)|^2 = \frac{\pi^4}9 + 8\sum_{n=1}^\infty \frac1{n^4},$$ so that $$\sum_{n=1}^\infty \frac1{n^4} = \frac18\left(\frac{\pi^4}5 - \frac{\pi^4}9\right) = \frac{\pi^4}{90}.$$ The sum of the even-numbered terms of that series is $$\displaystyle \sum_{n=1}^\infty \frac1{2^4n^4} = \frac1{16}\frac{\pi^4}{90}.$$ So the sum of the odd-numbered terms is $$\displaystyle \frac{15}{16}\frac{\pi^4}{90}$$. The alternating sum $$\displaystyle \sum_{n = 1}^\infty \frac{(-1)^{n-1}}{n^4}$$ is then the sum of the odd terms minus the sum of the even terms, namely $$\displaystyle \left(\frac{15}{16} - \frac1{16}\right)\frac{\pi^4}{90} = \frac{7\pi^4}{720}.$$

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