lim _n arrow ∞ 1/n [1/n sin^-1 1/n+2/n sin^-1 2/n+…+(π)/2 ]=

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MARKS Content Team · Accepted Answer

Correct answer: (C) (π)/8. Given that, lim _n arrow ∞ 1/n [1/n sin^-1 1/n+2/n sin^-1 2/n+…+(π)/2 ] The above expression may be written as =lim _n arrow ∞ frac∑_r=1^n r/n sin^-1 r/nn=∫_0^1 underset II x sin _I^-1 x d x Solving integration by parts method, we get =sin^-1 x ∫_0^1 x d x-∫_0^1 (d/(d x) (sin^-1 x ) ∫ x d x ) d x = [sin^-1 x · (x^2)/2 ]_0^1-∫_0^1 1/(√(1-x^2)) × (x^2)/2 d x = [sin^-1 x · (x^2)/2 ]_0^1-1/2 ∫_0^1 (x^2-1+1)/(√(1-x^2)) d x = [sin^-1(1) · 1/2-sin^-1(0) · (0^2)/2 ]-1/2 ∫_0^1 (- (1-x^2 ))/(√(1-x^2)) d x-1/2 ∫_0^1 (d x)/(√(1-x^2)) = [(π)/2 × 1/2-0 ]+1/2 ∫_0^1 √(1-x^2) d x-1/2 ∫_0^1 (d x)/(√(1-x^2)) =(π)/4+1/2 [x/2 √(1-x^2)+1/2 sin^-1 x ]_0^1-1/2 [sin^-1 x ]_0^1 [∵ ∫ √(a^2-x^2) d x=1/(2 a) √(a^2-x^2)+1/(2 a) sin^-1 (x/a ) . and .∫ (d x)/(√(a^2-x^2))=sin^-1 (x/a ) ] =(π)/4+1/2 [1/2 √(1-1^2)+1/2 sin^-1(1)-0/2 √(1-0^2) . .-0/2 sin^-1(0) (-1)/2 [sin^-1(2)-sin^-1(0) ] ] =(π)/4+1/2 [0+1/2 (π)/2-0-0 ]-1/2 [(π)/2-0 ] =(π)/4+1/2 [(π)/4-0 ]-(π)/4=(π)/4+(π)/8-(π)/4=(π)/8

$\lim _{n \rightarrow \infty} \frac{1}{n}\left[\frac{1}{n} \sin ^{-1} \frac{1}{n}+\frac{2}{n} \sin ^{-1} \frac{2}{n}+\ldots+\frac{\pi}{2}\right]=$

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