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Question Number 191168 by Mingma last updated on 19/Apr/23

Answered by mehdee42 last updated on 19/Apr/23

A=lim_(n→∞) ((((n+1)(n+2)...(2n)))^(1/n) /n)=lim_(n→∞) ((((n+1)(n+2)...(2n))/n^n ))^(1/n) ⇒lnA=lim_(n→∞) (1/n)[ln(1+(1/n))+ln(1+(2/n))+...+ln(1+(n/n))] =lim_(n→∞) (1/n)Σ_(i=1) ^n ln(1+(i/n))=∫^1 _0 ln(1+x)dx =[(1+x)ln(1+x)−x]_0 ^1 =2ln2−1=ln(4/e)⇒A=(4/e) ✓

A=limn→∞(n+1)(n+2)...(2n)nn=limn→∞(n+1)(n+2)...(2n)nnn⇒lnA=limn→∞1n[ln(1+1n)+ln(1+2n)+...+ln(1+nn)]=limn→∞1n∑ni=1ln(1+in)=∫01ln(1+x)dx=[(1+x)ln(1+x)−x]01=2ln2−1=ln4e⇒A=4e✓

Commented by Mingma last updated on 19/Apr/23

Perfect ��

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