Aufgaben:Problem 5

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Revision as of 17:24, 22 December 2014 by Nik (Talk | contribs) (Created page with "=Problem= Show, by using the Fourier series, that $$\sum_{k=1}^\infty \frac{1}{k^n} = - \frac{(2 \pi i)^n}{2(n!)} B_n,\ n \in 2 \mathbb{Z}_{>0}$$ where \(B_n\) are the Bernoul...")

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Problem

Show, by using the Fourier series, that $$\sum_{k=1}^\infty \frac{1}{k^n} = - \frac{(2 \pi i)^n}{2(n!)} B_n,\ n \in 2 \mathbb{Z}_{>0}$$ where \(B_n\) are the Bernoulli numbers defined as $$\frac{z}{e^z - 1} = \sum_{n=0}^\infty B_n \frac{z^n}{n!}$$

Hint: consider \( B_n(x) := \sum_{k \in \mathbb{Z} \setminus \{0\}} k^{-n} e^{ikx},\ n \in \mathbb{Z}_{\geq 2},\ x \in \mathbb{R} \)

Solution

Coming soon ;) Nik (talk) 18:24, 22 December 2014 (CET)