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lim-x-1-n-n-1-n-1-n-2-n-n-3-1-n-1-n-n-what-answer-




Question Number 129385 by Adel last updated on 15/Jan/21
lim_(x→∞) (1/( (√(n ))))( /( (√(n+1))))+( /( (√n)))(( 1)/( (√(n+2))))( / )+( /( (√n)))( /( (√(n+3))))(1/ )( / )+..............+( /( (√n)))(1/( (√(n+n))))( / )=?   what answer
$$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\frac{\mathrm{1}}{\:\sqrt{\mathrm{n}\:}}\frac{\:}{\:\sqrt{\mathrm{n}+\mathrm{1}}}+\frac{\:}{\:\sqrt{\mathrm{n}}}\frac{\:\mathrm{1}}{\:\sqrt{\mathrm{n}+\mathrm{2}}}\frac{\:}{\:}+\frac{\:}{\:\sqrt{\boldsymbol{\mathrm{n}}}}\frac{\:}{\:\sqrt{\boldsymbol{\mathrm{n}}+\mathrm{3}}}\frac{\mathrm{1}}{\:}\frac{\:}{\:}+…………..+\frac{\:}{\:\sqrt{\mathrm{n}}}\frac{\mathrm{1}}{\:\sqrt{\mathrm{n}+\mathrm{n}}}\frac{\:}{\:}=? \\ $$$$\:{what}\:{answer} \\ $$
Answered by Dwaipayan Shikari last updated on 15/Jan/21
lim_(n→∞) Σ_(k=1) ^n (1/( (√(n^2 +nk))))=(1/n)Σ_(k=1) ^n (1/( (√(1+(k/n)))))=∫_0 ^1 (1/( (√(1+x))))dx  =[2(√(1+x))]_0 ^1 =2(√2)−2
$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{\mathrm{1}}{\:\sqrt{{n}^{\mathrm{2}} +{nk}}}=\frac{\mathrm{1}}{{n}}\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}+\frac{{k}}{{n}}}}=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}}{\:\sqrt{\mathrm{1}+{x}}}{dx} \\ $$$$=\left[\mathrm{2}\sqrt{\mathrm{1}+{x}}\right]_{\mathrm{0}} ^{\mathrm{1}} =\mathrm{2}\sqrt{\mathrm{2}}−\mathrm{2} \\ $$
Commented by Adel last updated on 15/Jan/21
  Thank you, madam, what is the name of the method you have used and you have taken this antigral from zero to one?
$$ \\ $$Thank you, madam, what is the name of the method you have used and you have taken this antigral from zero to one?
Commented by Adel last updated on 16/Jan/21
answer   sir
$$\mathrm{answer}\:\:\:\mathrm{sir}\:\:\: \\ $$
Commented by Adel last updated on 16/Jan/21
name of the method?
$$\mathrm{name}\:\mathrm{of}\:\mathrm{the}\:\mathrm{method}? \\ $$
Commented by Adel last updated on 18/Jan/21
name the method plese sir
$$\mathrm{name}\:\mathrm{the}\:\mathrm{method}\:\mathrm{plese}\:\mathrm{sir} \\ $$
Commented by Dwaipayan Shikari last updated on 22/Jan/21
Riemann sums  You can Google it
$${Riemann}\:{sums} \\ $$$${You}\:{can}\:{Google}\:{it} \\ $$

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