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) ✓](Q191180.png)
$${A}={lim}_{{n}\rightarrow\infty} \frac{\sqrt[{{n}}]{\left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right)...\left(\mathrm{2}{n}\right)}}{{n}}={lim}_{{n}\rightarrow\infty} \sqrt[{{n}}]{\frac{\left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right)...\left(\mathrm{2}{n}\right)}{{n}^{{n}} }} \\ $$$$\Rightarrow{lnA}={lim}_{{n}\rightarrow\infty} \:\frac{\mathrm{1}}{{n}}\left[{ln}\left(\mathrm{1}+\frac{\mathrm{1}}{{n}}\right)+{ln}\left(\mathrm{1}+\frac{\mathrm{2}}{{n}}\right)+...+{ln}\left(\mathrm{1}+\frac{{n}}{{n}}\right)\right] \\ $$$$={lim}_{{n}\rightarrow\infty} \:\frac{\mathrm{1}}{{n}}\underset{{i}=\mathrm{1}} {\overset{{n}} {\sum}}{ln}\left(\mathrm{1}+\frac{{i}}{{n}}\right)=\underset{\mathrm{0}} {\int}^{\mathrm{1}} {ln}\left(\mathrm{1}+{x}\right){dx} \\ $$$$=\left[\left(\mathrm{1}+{x}\right){ln}\left(\mathrm{1}+{x}\right)−{x}\right]_{\mathrm{0}} ^{\mathrm{1}} =\mathrm{2}{ln}\mathrm{2}−\mathrm{1}={ln}\frac{\mathrm{4}}{{e}}\Rightarrow{A}=\frac{\mathrm{4}}{{e}}\:\checkmark \\ $$
Commented by Mingma last updated on 19/Apr/23
Perfect ��