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Question Number 150226 by mnjuly1970 last updated on 10/Aug/21
       prove  that ::                                  ζ (0 )=^?  ((−1)/2) ..........■         m.n...
$$ \\ $$$$\:\:\:\:\:\mathrm{prove}\:\:\mathrm{that}\::: \\ $$$$\:\:\:\:\:\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\zeta\:\left(\mathrm{0}\:\right)\overset{?} {=}\:\frac{−\mathrm{1}}{\mathrm{2}}\:……….\blacksquare \\ $$$$\:\:\:\:\:\:\:{m}.{n}… \\ $$
Answered by Kamel last updated on 10/Aug/21
By using fonctionnal relation of zeta.
$${By}\:{using}\:{fonctionnal}\:{relation}\:{of}\:{zeta}. \\ $$
Commented by mnjuly1970 last updated on 10/Aug/21
  yes mr kamel....mercey...
$$\:\:{yes}\:{mr}\:{kamel}….{mercey}… \\ $$
Answered by Kamel last updated on 10/Aug/21
ζ(2n)=(((−1)^(n−1) B_(2n) (2π)^(2n) )/(2(2n)!)), B_n  denote Bernoulli number′s.  ∴  ζ(0)=−(B_0 /2)=−(1/2)
$$\zeta\left(\mathrm{2}{n}\right)=\frac{\left(−\mathrm{1}\right)^{{n}−\mathrm{1}} {B}_{\mathrm{2}{n}} \left(\mathrm{2}\pi\right)^{\mathrm{2}{n}} }{\mathrm{2}\left(\mathrm{2}{n}\right)!},\:{B}_{{n}} \:{denote}\:{Bernoulli}\:{number}'{s}. \\ $$$$\therefore\:\:\zeta\left(\mathrm{0}\right)=−\frac{{B}_{\mathrm{0}} }{\mathrm{2}}=−\frac{\mathrm{1}}{\mathrm{2}} \\ $$

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