Question Number 111766 by mathmax by abdo last updated on 04/Sep/20
$$\mathrm{calculate}\:\mathrm{lim}_{\mathrm{n}\rightarrow+\infty} \:\sum_{\mathrm{k}=\mathrm{1}} ^{\mathrm{n}} \mathrm{ln}\left(\frac{\mathrm{n}}{\mathrm{n}+\mathrm{k}}\right)^{\frac{\mathrm{1}}{\mathrm{n}}} \\ $$
Answered by Dwaipayan Shikari last updated on 04/Sep/20
![lim_(n→∞) (1/n)Σ^n log((1/(1+(k/n))))=∫_0 ^1 log((1/(1+x)))=−[xlog(1+x)]_0 ^1 +∫_0 ^1 (x/(1+x)) =−log(2)+1−[log(1+x)]_0 ^1 =1−2log(2)](https://www.tinkutara.com/question/Q111775.png)
$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\frac{\mathrm{1}}{{n}}\overset{{n}} {\sum}{log}\left(\frac{\mathrm{1}}{\mathrm{1}+\frac{{k}}{{n}}}\right)=\int_{\mathrm{0}} ^{\mathrm{1}} {log}\left(\frac{\mathrm{1}}{\mathrm{1}+{x}}\right)=−\left[{xlog}\left(\mathrm{1}+{x}\right)\right]_{\mathrm{0}} ^{\mathrm{1}} +\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{x}}{\mathrm{1}+{x}} \\ $$$$=−{log}\left(\mathrm{2}\right)+\mathrm{1}−\left[{log}\left(\mathrm{1}+{x}\right)\right]_{\mathrm{0}} ^{\mathrm{1}} =\mathrm{1}−\mathrm{2}{log}\left(\mathrm{2}\right) \\ $$
Answered by mathmax by abdo last updated on 07/Sep/20
![A_n =(1/n)Σ_(k=1) ^n ln((n/(n+k))) =(1/n) Σ_(k=1) ^n ln((1/(1+(k/n)))) ⇒A_n is a Rieman sum lim_(n→+∞) A_n =∫_0 ^1 ln((1/(1+x)))dx =−∫_0 ^1 ln(1+x)dx =_(1+x=t) =−∫_1 ^2 ln(t)dt =−[tlnt−t]_1 ^2 =−{2ln2−2+1} =−{2ln2−1} =1−2ln(2)](https://www.tinkutara.com/question/Q112253.png)
$$\mathrm{A}_{\mathrm{n}} =\frac{\mathrm{1}}{\mathrm{n}}\sum_{\mathrm{k}=\mathrm{1}} ^{\mathrm{n}} \:\mathrm{ln}\left(\frac{\mathrm{n}}{\mathrm{n}+\mathrm{k}}\right)\:=\frac{\mathrm{1}}{\mathrm{n}}\:\sum_{\mathrm{k}=\mathrm{1}} ^{\mathrm{n}} \:\mathrm{ln}\left(\frac{\mathrm{1}}{\mathrm{1}+\frac{\mathrm{k}}{\mathrm{n}}}\right)\:\Rightarrow\mathrm{A}_{\mathrm{n}} \mathrm{is}\:\:\mathrm{a}\:\mathrm{Rieman}\:\mathrm{sum} \\ $$$$\mathrm{lim}_{\mathrm{n}\rightarrow+\infty} \mathrm{A}_{\mathrm{n}} =\int_{\mathrm{0}} ^{\mathrm{1}} \:\mathrm{ln}\left(\frac{\mathrm{1}}{\mathrm{1}+\mathrm{x}}\right)\mathrm{dx}\:=−\int_{\mathrm{0}} ^{\mathrm{1}} \:\mathrm{ln}\left(\mathrm{1}+\mathrm{x}\right)\mathrm{dx}\:=_{\mathrm{1}+\mathrm{x}=\mathrm{t}} \\ $$$$=−\int_{\mathrm{1}} ^{\mathrm{2}} \:\mathrm{ln}\left(\mathrm{t}\right)\mathrm{dt}\:=−\left[\mathrm{tlnt}−\mathrm{t}\right]_{\mathrm{1}} ^{\mathrm{2}} \:=−\left\{\mathrm{2ln2}−\mathrm{2}+\mathrm{1}\right\}\:=−\left\{\mathrm{2ln2}−\mathrm{1}\right\} \\ $$$$=\mathrm{1}−\mathrm{2ln}\left(\mathrm{2}\right) \\ $$