The series is given: \begin{equation} \sum_{n=1}^\infty \frac{1-\cos(nx)}{n^4} = \tfrac{1}{12}x^2 (\pi - \tfrac{1}{2}x)^2 \end{equation} My first attempt to prove this equality was to use Mellin transform, but I fail to show the result.
Asked
Active
Viewed 78 times
0
-
3Why don't you simply compute the Fourier series of the right hand side (when considered as periodic function which is what the equality is meant to be) – lcv Jul 26 '22 at 10:03
-
1The formula holds for $x\in [0, 2\pi]$, e.g., $\mathrm{LHS} \ne \mathrm{RHS}$ for $x = -1$. – River Li Jul 26 '22 at 12:37
1 Answers
1
By using the series \begin{align} \sum_{n=1}^{\infty} \frac{ \cos(nx) }{ n^{2} } = \zeta(2) - \frac{ \pi \, x}{2} + \frac{x^{2}}{4}\tag 1 \end{align} integrate to obtain \begin{align} \sum_{n=1}^{\infty} \frac{\sin(n x)}{n^{3}} = \zeta(2) \, x - \frac{\pi \, x^{2}}{4} + \frac{x^{3}}{12}. \end{align} Integrate once again so we get \begin{align} \sum_{n=1}^{\infty} \frac{\cos(nx)}{n^{4}} = \zeta(4) - \frac{\zeta(2) \, x^{2}}{2} + \frac{\pi \, x^{3}}{12} - \frac{x^{4}}{48}. \end{align} So: $$\sum_{n=1}^\infty \frac{1-\cos(nx)}{n^4} =\sum_{n=1}^\infty\frac{1}{n^4}-\sum_{n=1}^\infty\frac{\cos(nx)}{n^4}=\\\zeta(4)-\zeta(4)+\frac{\zeta(2)x^2}{2}-\frac{\pi x^3}{12}+\frac{x^4}{48}=\frac{\pi^2 x^2}{12}-\frac{\pi x^3}{12}+\frac{x^4}{48}=\frac{x^2}{12}(\pi-\frac{x}{2})^2$$
To prove the equality $(1)$ you can check here Series $\sum_{n=1}^{\infty}\frac{\cos(nx)}{n^2}$
Gary
- 31,845
-
-
-
-
-
1$\int_{0}^{x}\sum\frac{\sin (nt)}{n^3};dt=\sum\frac{\cos (nx)}{n^4}-\sum\frac{\cos (0)}{n^4}$, and $\sum\frac{\cos (0)}{n^4}=\sum\frac{1}{n^4}=\zeta(4)$ – Guillermo García Sáez Jul 26 '22 at 10:59
-
-
-
As River Li mentions in the comments, the formula holds for $x\in [0,2π]$. – Bjaam Jul 26 '22 at 13:15