Question Number 116317 by bemath last updated on 03/Oct/20
$$\:\:\underset{\mathrm{n}\rightarrow\infty} {\mathrm{lim}}\:\left(\frac{\sqrt{\mathrm{n}+\mathrm{1}}+\sqrt{\mathrm{n}+\mathrm{2}}+\sqrt{\mathrm{n}+\mathrm{3}}+…+\sqrt{\mathrm{2n}−\mathrm{1}}}{\mathrm{n}^{\frac{\mathrm{3}}{\mathrm{2}}} }\:\right)\:= \\ $$
Answered by Bird last updated on 03/Oct/20
![U_n =(1/n^(3/2) )Σ_(k=1) ^(n−1) (√(n+k)) =(1/n^(3/2) ).(√n)Σ_(k=1) ^(n−1) (√(1+(k/n))) =(1/n^((3/2)−(1/2)) )Σ_(k=1) ^(n−1) (√(1+(k/n))) =(1/n)Σ_(k=1) ^(n−1) (√(1+(k/n)))→∫_0 ^1 (√(1+x))dx ∫_0 ^1 (√(1+x))dx =_((√(1+x))=t) ∫_1 ^(√2) t(2t)dt =2 ∫_1 ^(√2) t^(2 ) dt =(2/3)[t^3 ]_1 ^(√2) =(2/3){3(√2)−1} =2(√2)−(2/3)](https://www.tinkutara.com/question/Q116320.png)
$${U}_{{n}} \:=\frac{\mathrm{1}}{{n}^{\frac{\mathrm{3}}{\mathrm{2}}} }\sum_{{k}=\mathrm{1}} ^{{n}−\mathrm{1}} \sqrt{{n}+{k}} \\ $$$$=\frac{\mathrm{1}}{{n}^{\frac{\mathrm{3}}{\mathrm{2}}} }.\sqrt{{n}}\sum_{{k}=\mathrm{1}} ^{{n}−\mathrm{1}} \sqrt{\mathrm{1}+\frac{{k}}{{n}}} \\ $$$$=\frac{\mathrm{1}}{{n}^{\frac{\mathrm{3}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{2}}} }\sum_{{k}=\mathrm{1}} ^{{n}−\mathrm{1}} \sqrt{\mathrm{1}+\frac{{k}}{{n}}} \\ $$$$=\frac{\mathrm{1}}{{n}}\sum_{{k}=\mathrm{1}} ^{{n}−\mathrm{1}} \sqrt{\mathrm{1}+\frac{{k}}{{n}}}\rightarrow\int_{\mathrm{0}} ^{\mathrm{1}} \sqrt{\mathrm{1}+{x}}{dx} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \sqrt{\mathrm{1}+{x}}{dx}\:=_{\sqrt{\mathrm{1}+{x}}={t}} \:\:\int_{\mathrm{1}} ^{\sqrt{\mathrm{2}}} {t}\left(\mathrm{2}{t}\right){dt} \\ $$$$=\mathrm{2}\:\int_{\mathrm{1}} ^{\sqrt{\mathrm{2}}} {t}^{\mathrm{2}\:} {dt}\:=\frac{\mathrm{2}}{\mathrm{3}}\left[{t}^{\mathrm{3}} \right]_{\mathrm{1}} ^{\sqrt{\mathrm{2}}} \\ $$$$=\frac{\mathrm{2}}{\mathrm{3}}\left\{\mathrm{3}\sqrt{\mathrm{2}}−\mathrm{1}\right\}\:=\mathrm{2}\sqrt{\mathrm{2}}−\frac{\mathrm{2}}{\mathrm{3}} \\ $$$$ \\ $$
Commented by john santu last updated on 03/Oct/20
$${typo}\:=\:\frac{\mathrm{2}}{\mathrm{3}}\left(\mathrm{2}\sqrt{\mathrm{2}}−\mathrm{1}\right) \\ $$
Commented by Bird last updated on 03/Oct/20
$${yes}\:{thanks}\:{john} \\ $$
Answered by Dwaipayan Shikari last updated on 03/Oct/20
![(1/n)lim_(n→∞) ((√(1+(1/n)))+(√(1+(2/n))) +.....) lim_(n→∞) (1/n)Σ_(k=1) ^n ((√(1+(k/n)))) ∫_0 ^1 (√(1+x)) dx =(2/3)[(1+x)^(3/2) ]_0 ^1 =(2/3)(2(√2)−1)](https://www.tinkutara.com/question/Q116336.png)
$$\frac{\mathrm{1}}{\mathrm{n}}\underset{\mathrm{n}\rightarrow\infty} {\mathrm{lim}}\left(\sqrt{\mathrm{1}+\frac{\mathrm{1}}{\mathrm{n}}}+\sqrt{\mathrm{1}+\frac{\mathrm{2}}{\mathrm{n}}}\:\:+…..\right) \\ $$$$\underset{\mathrm{n}\rightarrow\infty} {\mathrm{lim}}\frac{\mathrm{1}}{\mathrm{n}}\underset{\mathrm{k}=\mathrm{1}} {\overset{\mathrm{n}} {\sum}}\left(\sqrt{\mathrm{1}+\frac{\mathrm{k}}{\mathrm{n}}}\right) \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \sqrt{\mathrm{1}+\mathrm{x}}\:\mathrm{dx} \\ $$$$=\frac{\mathrm{2}}{\mathrm{3}}\left[\left(\mathrm{1}+\mathrm{x}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} \right]_{\mathrm{0}} ^{\mathrm{1}} =\frac{\mathrm{2}}{\mathrm{3}}\left(\mathrm{2}\sqrt{\mathrm{2}}−\mathrm{1}\right) \\ $$