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Δημοσιεύτηκε: **Παρ Νοέμ 14, 2025 11:07 pm**

Για $k>\frac{1}{2}$ , να δειχθεί ότι:

$\displaystyle{\sum_{m=1}^{\infty} \sum_{n=1}^{\infty} \frac{(-1)^n}{n^2}\sin \frac{n}{m^{2k}}=\frac{\zeta(6k)}{12}-\frac{\pi^2}{12}\zeta(2k)}$

Δημοσιεύτηκε: **Κυρ Φεβ 22, 2026 1:08 pm**

Tolaso J Kos έγραψε: Παρ Νοέμ 14, 2025 11:07 pm Για $k>\frac{1}{2}$ , να δειχθεί ότι:

$\displaystyle{\sum_{m=1}^{\infty} \sum_{n=1}^{\infty} \frac{(-1)^n}{n^2}\sin \frac{n}{m^{2k}}=\frac{\zeta(6k)}{12}-\frac{\pi^2}{12}\zeta(2k)}$

$\displaystyle S = \sum_{m=1}^{\infty} \sum_{n=1}^{\infty} \frac{(-1)^n}{n^{\color{red}3}} \sin\left(\frac{n}{m^{2k}}\right)$

$\displaystyle \sum_{n=1}^{\infty} \frac{(-1)^n}{n^2} \cos(nt) = \frac{t^2}{4} - \frac{\pi^2}{12} \,\, , \,\,\,\,\,\, \left| t \right| \leq \pi \,\,\,\,(\text{Fourier Series})$

$\displaystyle \Rightarrow \int_{0}^{x} \sum_{n=1}^{\infty} \frac{(-1)^n}{n^2} \cos(nt) \, dt = \int_{0}^{x} \left( \frac{t^2}{4} - \frac{\pi^2}{12} \right) \, dt \Rightarrow \sum_{n=1}^{\infty} \frac{(-1)^n}{n^3} \sin(nx) = \frac{x^3 - \pi^2 x}{12}$

$\displaystyle \overset{x = m^{-2k}}{\underset{k>1/2}{\Longrightarrow}} \,\,\sum_{n=1}^{\infty} \frac{(-1)^n}{n^3} \sin\left(\frac{n}{m^{2k}}\right) = \frac{1}{12 m^{6k}} - \frac{\pi^2}{12 m^{2k}}$

$\displaystyle \Rightarrow S = \sum_{m=1}^{\infty} \left( \frac{1}{12 m^{6k}} - \frac{\pi^2}{12 m^{2k}} \right) = \frac{1}{12} \sum_{m=1}^{\infty} \frac{1}{m^{6k}} - \frac{\pi^2}{12} \sum_{m=1}^{\infty} \frac{1}{m^{2k}} = \frac{\zeta(6k)}{12} - \frac{\pi^2}{12} \zeta(2k)$

https://math.stackexchange.com/question ... nn3-sinn-m

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