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Question Number 2777 by Filup last updated on 27/Nov/15
Prove that:  ζ(−2n)=0
$$\mathrm{Prove}\:\mathrm{that}: \\ $$$$\zeta\left(−\mathrm{2}{n}\right)=\mathrm{0} \\ $$
Answered by prakash jain last updated on 27/Nov/15
Functional Equation for ζ(s)  ζ(s)=2^s π^(s−1) sin (((πs)/2))Γ(1−s)ζ(1−s)  s=−2n, n∈N^∗   ζ(−2n)=0
$$\mathrm{Functional}\:\mathrm{Equation}\:\mathrm{for}\:\zeta\left({s}\right) \\ $$$$\zeta\left({s}\right)=\mathrm{2}^{{s}} \pi^{{s}−\mathrm{1}} \mathrm{sin}\:\left(\frac{\pi{s}}{\mathrm{2}}\right)\Gamma\left(\mathrm{1}−{s}\right)\zeta\left(\mathrm{1}−{s}\right) \\ $$$${s}=−\mathrm{2}{n},\:{n}\in\mathbb{N}^{\ast} \\ $$$$\zeta\left(−\mathrm{2}{n}\right)=\mathrm{0} \\ $$
Commented by 123456 last updated on 27/Nov/15
n∈N^∗
$${n}\in\mathbb{N}^{\ast} \\ $$
Commented by prakash jain last updated on 27/Nov/15
corrected.
$$\mathrm{corrected}. \\ $$

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