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From: Olivier Miakinen <om+news@miakinen.net>
Newsgroups: fr.sci.maths
Subject: Re: lim n sin(2pi exp(1) n!) ?
Date: Wed, 18 Aug 2021 15:58:17 +0200
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Bonjour,

Le 18/08/2021 à 15:24, Samuel DEVULDER a écrit :
> 
> Sauriez vous calculer:
> 
> |        lim   n sin(2pi exp(1) n!)
> |       n->oo
> 
> En déduire que exp(1) est irrationnel.

Yn cerzvèer cnegvr zr frzoyr nffrm qvssvpvyr, nybef wr invf pbzzrapre
cne yn qrhkvèzr.

Fhccbfbaf dhr rkc(1) fbvg engvbaary. Ba crhg nybef y'épever rkc(1)=n/o.
Ba n nybef rkc(1)×o! = n×(o-1)! dhv rfg ha ragvre, rg cbhe gbhg a > o
rkc(1)×a! = n×(o-1)!×(o+1)×...×a dhv rfg nhffv ha ragvre. Cne pbafédhrag
à cnegve qr a=o ba n gbhwbhef a fva(2cv rkc(1) a!) = 0, rg yn yvzvgr
inhg 0.

Ra pbapyhfvba, fv ba cebhir qnaf yn cerzvèer cnegvr dhr prggr yvzvgr rfg
qvsséeragr qr 0, pryn cebhiren dhr rkc(1) rfg veengvbaary.


-- 
Olivier Miakinen