Question Number 100904 by Dwaipayan Shikari last updated on 29/Jun/20

lim_(n→∞) [(((n+1)(n+2)......3n)/n^(2n) )]^(1/n)

Commented byAr Brandon last updated on 29/Jun/20

I like how you transformed the sum directly into integral without decomposing it into simpler sums.😃

Commented byDwaipayan Shikari last updated on 29/Jun/20

suppose y=lim_(n→∞) [(((n+1)(n+2)....3n)/(n.n......2n times))]^(1/n) log y=(1/n) lim_(n→∞) [log(1+(1/n))+log(1+(2/n))+....+log(1+((2n)/n))] logy=(1/n) lim_(n→∞ ) Σ_(r=1) ^(2n) log(1+(r/n)) logy=∫_0 ^2 log(1+x)dx logy= [xlog(1+x)]_0 ^2 −∫_0 ^2 (x/(x+1)) logy= 2log3−[x−log(x+1)]_0 ^2 logy=2log3−2+log3 logy=3log3−2 y= e^(3log3−2) =((27)/e^2 ) check this

Commented bymscbed last updated on 29/Jun/20

Answered by Ar Brandon last updated on 29/Jun/20

l=lim_(n→∞) [(((n+1)(n+2)...(n+2n))/n^(2n) )]^(1/n) =lim_(n→∞) [(1+(1/n))(1+(2/n))∙∙∙(1+((2n)/n))]^(1/n) ⇒l=lim_(n→∞) Π_(k=1) ^(2n) (1+(k/n))^(1/n) ⇒lnl=lim_(n→∞) ln{Π_(k=1) ^(2n) (1+(k/n))^(1/n) } ⇒lnl=lim_(n→∞) Σ_(k=1) ^(2n) {ln(1+(k/n))^(1/n) }=lim_(n→∞) (1/n)Σ_(k=1) ^(2n) ln(1+(k/n)) =lim_(n→∞) (1/n){Σ_(k=1) ^n ln(1+(k/n))+Σ_(k=n+1) ^(2n) ln(1+(k/n))} =lim_(n→∞) (1/n){Σ_(k=1) ^n ln(1+(k/n))+Σ_(p=1) ^n ln(1+((n+p)/n))} =lim_(n→∞) (1/n)Σ_(k=1) ^n ln(1+(k/n))+lim_(n→∞) (1/n)Σ_(p=1) ^n ln(2+(p/n)) =∫_1 ^2 lnxdx+∫_2 ^3 lnxdx=[x(lnx−1)]_1 ^2 +[x(lnx−1)]_2 ^3 =2(ln2−1)+1+3(ln3−1)−2(ln2−1)=3ln3−2 ⇒l=e^(3ln3−2) =(3^3 /e^2 )⇒lim_(n→∞) [(((n+1)(n+2)...(n+2n))/n^(2n) )]^(1/n) =((27)/e^2 )

Answered by mathmax by abdo last updated on 29/Jun/20

let A_n ={(((n+1)(n+2)....(n+2n))/n^(2n) )}^(1/n) ⇒lnA_n =(1/n)ln(((Π_(k=1) ^(2n) (n+k))/n^(2n) )) =(1/n){Σ_(k=1) ^(2n) ln(n+k)−2nln(n)}=(1/n)Σ_(k=1) ^(2n) ln (n+k)−2ln(n) =(1/n)Σ_(k=1) ^(2n) ( lnn+ln(1+(k/n)))−2ln(n) =(1/n)Σ_(k=1) ^(2n) ln(1+(k/n)) =(1/n)Σ_(k=1) ^(n−1) ln(1+(k/n)) +(1/n)Σ_(k=n) ^(2n) ln(1+(k/n))(→p=k−n) =(1/n)Σ_(k=1) ^(n−1) ln(1+(k/n)) +(1/n)Σ_(p=0) ^n ln(1+((p+n)/n)) =(1/n)Σ_(k=1) ^(n−1) ln(1+(k/n))+(1/n)Σ_(p=0) ^n ln(2+(p/n)) we have (1/n)Σ_(k=1) ^(n−1) ln(1+(k/n))→∫_0 ^1 ln(1+x)dx =_(1+x=t) ∫_1 ^2 ln(t)dt =[tlnt−t]_1 ^2 =2ln(2)−2+1 =2ln(2)−1 (1/n)Σ_(p=0) ^n ln(2+(p/n))→∫_0 ^1 ln(2+x)dx =_(2+x=t) ∫_2 ^3 ln(t)dt=[tlnt−t]_2 ^3 =3ln(3)−3−2ln(2)+2 =3ln(3)−2ln(2)−1 ⇒ lim_(n→+∞) lnA_n =2ln(2)−1+3ln(3)−2ln(2)−1 =3ln(3)−2 ⇒ lim_(n→+∞) A_n =e^(3ln(3)−2) =9e^(−2) =(9/e^2 )

Commented bymathmax by abdo last updated on 29/Jun/20

sorry lim_(n→+∞) A_n =27 e^(−2) =((27)/e^2 )

Answered by Ar Brandon last updated on 29/Jun/20

A_n =lim_(n→∞) [(((n+1)(n+2)...(n+2n))/n^(2n) )]^(1/n) =lim_(n→∞) [(1+(1/n))(1+(2/n))∙∙∙(1+((2n)/n))]^(1/n) ⇒A_n =lim_(n→∞) Π_(k=1) ^(2n) (1+(k/n))^(1/n) ⇒lnA_n =lim_(n→∞) ln{Π_(k=1) ^(2n) (1+(k/n))^(1/n) } ⇒lnA_n =lim_(n→∞) Σ_(k=1) ^(2n) {ln(1+(k/n))^(1/n) }=lim_(n→∞) (1/n)Σ_(k=1) ^(2n) ln(1+(k/n)) =∫_1 ^3 lnxdx=[x(lnx−1)]_1 ^3 =3(ln3−1)−(ln1−1) =3ln3−3+1=3ln3−2⇒l=e^(3ln3−2) =(3^3 /e^2 ) ⇒lim_(n→∞) [(((n+1)(n+2)...(n+2n))/n^(2n) )]^(1/n) =((27)/e^2 )