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Question-149180




Question Number 149180 by mathdanisur last updated on 03/Aug/21
Answered by Kamel last updated on 03/Aug/21
L=lim_(n→+∞) e^(nLn(Σ_(k=1) ^n (1/( (√(n^2 +k)))))) =lim_(n→+∞) e^(nLn((1/n)Σ_(k=1) ^n (1/( (√(1+(k/n^2 )))))))   (1/( (√(1+(k/n^2 )))))∼1−(k/(2n^2 ))+o((1/n^2 )) , (Σ_(k=1) ^n (k^m /n^(2m) )→0,n→+∞)  ∴ L=lim_(n→+∞) e^(nLn((1/n)(n−((n(n+1))/(4n^2 )))) =lim_(n→+∞) e^(n(Ln(1−(1/(4n))))) =e^(−(1/4))
$${L}=\underset{{n}\rightarrow+\infty} {{lim}e}^{{nLn}\left(\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{\mathrm{1}}{\:\sqrt{{n}^{\mathrm{2}} +{k}}}\right)} =\underset{{n}\rightarrow+\infty} {{lim}e}^{{nLn}\left(\frac{\mathrm{1}}{{n}}\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}+\frac{{k}}{{n}^{\mathrm{2}} }}}\right)} \\ $$$$\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}+\frac{{k}}{{n}^{\mathrm{2}} }}}\sim\mathrm{1}−\frac{{k}}{\mathrm{2}{n}^{\mathrm{2}} }+{o}\left(\frac{\mathrm{1}}{{n}^{\mathrm{2}} }\right)\:,\:\left(\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{{k}^{{m}} }{{n}^{\mathrm{2}{m}} }\rightarrow\mathrm{0},{n}\rightarrow+\infty\right) \\ $$$$\therefore\:{L}=\underset{{n}\rightarrow+\infty} {{lim}e}^{{nLn}\left(\frac{\mathrm{1}}{{n}}\left({n}−\frac{{n}\left({n}+\mathrm{1}\right)}{\mathrm{4}{n}^{\mathrm{2}} }\right)\right.} =\underset{{n}\rightarrow+\infty} {{lim}e}^{{n}\left({Ln}\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{4}{n}}\right)\right)} ={e}^{−\frac{\mathrm{1}}{\mathrm{4}}} \\ $$
Commented by mathdanisur last updated on 03/Aug/21
Thank You Ser
$${Thank}\:{You}\:{Ser} \\ $$

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