Question Number 30757 by abdo imad last updated on 25/Feb/18
$${find}\:{lim}_{{n}\rightarrow\infty} \:\:\left(\frac{\mathrm{1}}{{n}}\:+\frac{\mathrm{1}}{\:\sqrt{{n}^{\mathrm{2}} \:−\mathrm{1}}}\:+….\:+\frac{\mathrm{1}}{\:\sqrt{{n}^{\mathrm{2}} \:−\left({n}−\mathrm{1}\right)^{\mathrm{2}} }}\:\right) \\ $$
Commented by abdo imad last updated on 28/Feb/18
![let put w_n = (1/n) +(1/( (√(n^2 −1)))) +...+(1/( (√(n^2 −(n−1)^2 )))) w_n =(1/n)( 1+(1/( (√(1−(1^2 /n^2 ))))) +(1/( (√(1−(2^2 /n^2 ))))) +..... + (1/( (√(1−(((n−1)^2 )/n^2 )))))) = (1/n)Σ_(k=0) ^(n−1) (1/( (√(1 −((k/n))^2 )))) so w_n is a Rieman sum and lim_(n→∞) w_n = ∫_0 ^1 (dx/( (√(1−x^2 )))) =[arcsinx]_0 ^1 =(π/2) .](https://www.tinkutara.com/question/Q30906.png)
$${let}\:{put}\:{w}_{{n}} =\:\frac{\mathrm{1}}{{n}}\:+\frac{\mathrm{1}}{\:\sqrt{{n}^{\mathrm{2}} −\mathrm{1}}}\:+…+\frac{\mathrm{1}}{\:\sqrt{{n}^{\mathrm{2}} \:−\left({n}−\mathrm{1}\right)^{\mathrm{2}} }} \\ $$$${w}_{{n}} =\frac{\mathrm{1}}{{n}}\left(\:\mathrm{1}+\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}−\frac{\mathrm{1}^{\mathrm{2}} }{{n}^{\mathrm{2}} }}}\:+\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}−\frac{\mathrm{2}^{\mathrm{2}} }{{n}^{\mathrm{2}} }}}\:+…..\:+\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}−\frac{\left({n}−\mathrm{1}\right)^{\mathrm{2}} }{{n}^{\mathrm{2}} }}}\right) \\ $$$$=\:\frac{\mathrm{1}}{{n}}\sum_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \:\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}\:−\left(\frac{{k}}{{n}}\right)^{\mathrm{2}} }}\:{so}\:{w}_{{n}} \:{is}\:{a}\:{Rieman}\:{sum}\:{and} \\ $$$${lim}_{{n}\rightarrow\infty} \:{w}_{{n}} \:\:=\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\:\frac{{dx}}{\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}\:=\left[{arcsinx}\right]_{\mathrm{0}} ^{\mathrm{1}} \:=\frac{\pi}{\mathrm{2}}\:. \\ $$