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Question Number 144483 by alcohol last updated on 25/Jun/21

(p_n )=(1+(1/n^2 ))(1+(2/n^2 ))...(1+(n/n^2 ))  Σ_(k=1) ^n k^2 =(1/6)n(2n+1)(n+1)  show that  (1/2)(1+(1/n))−(1/(12n^2 ))(2n+1)(n+1)<ln(p_n )<(1/2)(1+(1/n))  hence find lim_(n→∞) (p_n )  2) show that   t−(t^2 /2) ≤ln(1+t) ≤t, ∀t>0  please help

(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 tt22ln(1+t)t,t>0 pleasehelp

Answered by Olaf_Thorendsen last updated on 26/Jun/21

ln(1+x) = Σ_(n=0) ^∞ (−1)^n (x^(n+1) /(n+1))  ⇒ x−(x^2 /2) < ln(1+x) < x    (1)  p_n  = Π_(k=1) ^n (1+(k/n^2 ))  lnp_n  = Σ_(k=1) ^n ln(1+(k/n^2 ))  (1) : (k/n^2 )−(k^2 /(2n^4 )) < ln(1+(k/n^2 )) < (k/n^2 )  Σ_(k=1) ^n ((k/n^2 )−(k^2 /(2n^4 ))) < lnp_n =Σ_(k=1) ^n ln(1+(k/n^2 )) <Σ_(k=1) ^n (k/n^2 )  ((n(n+1))/(2n^2 ))−((n(n+1)(2n+1))/(12n^4 )) < p_n  < ((n(n+1))/(2n^2 ))  (1/2)(1+(1/n))−(((n+1)(2n+1))/(12n^3 )) < p_n  < (1/2)(1+(1/n))

ln(1+x)=n=0(1)nxn+1n+1 xx22<ln(1+x)<x(1) pn=nk=1(1+kn2) lnpn=nk=1ln(1+kn2) (1):kn2k22n4<ln(1+kn2)<kn2 nk=1(kn2k22n4)<lnpn=nk=1ln(1+kn2)<nk=1kn2 n(n+1)2n2n(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

thank you brother

thankyoubrother

Commented bypuissant last updated on 07/Jul/21

Desole mais il ya erreur c′est ln(p_n ) qui  est encadre^� ..  ainsi  lim_(n→+∞) ln(p_n )=(1/2)  donc  lim_(n→+∞) (p_n )=(√e).

Desolemaisilyaerreurcestln(pn)qui estencadre´.. ainsilimn+ln(pn)=12 donclimn+(pn)=e.

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