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Question Number 144483 by alcohol last updated on 25/Jun/21
(pn)=(1+1n2)(1+2n2)...(1+nn2) ∑nk=1k2=16n(2n+1)(n+1) showthat 12(1+1n)−112n2(2n+1)(n+1)<ln(pn)<12(1+1n) hencefindlimn→∞(pn) 2)showthat t−t22⩽ln(1+t)⩽t,∀t>0 pleasehelp
Answered by Olaf_Thorendsen last updated on 26/Jun/21
ln(1+x)=∑∞n=0(−1)nxn+1n+1 ⇒x−x22<ln(1+x)<x(1) pn=∏nk=1(1+kn2) lnpn=∑nk=1ln(1+kn2) (1):kn2−k22n4<ln(1+kn2)<kn2 ∑nk=1(kn2−k22n4)<lnpn=∑nk=1ln(1+kn2)<∑nk=1kn2 n(n+1)2n2−n(n+1)(2n+1)12n4<pn<n(n+1)2n2 12(1+1n)−(n+1)(2n+1)12n3<pn<12(1+1n)
Commented byalcohol last updated on 26/Jun/21
thankyoubrother
Commented bypuissant last updated on 07/Jul/21
Desolemaisilyaerreurc′estln(pn)qui estencadre´.. ainsilimn→+∞ln(pn)=12 donclimn→+∞(pn)=e.
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