find lim_(n→+∞) (1/n^3 ) Σ_(k=1) ^n k^2 sin(((kπ)/n))


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Question Number 34692 by math khazana by abdo last updated on 09/May/18

$f i n d {l i m}_{n \to + \infty} \frac{1}{n^{3}} \sum_{k = 1}^{n} k^{2} s i n (\frac{k π}{n})$
Commented by math khazana by abdo last updated on 11/May/18
$let put S_n = (1/n^3 ) Σ_(k=1) ^n k^2 sin(((kπ)/n)) S_n = (1/n) Σ_(k=1) ^n ((k/n))^2 sin(((kπ)/n))→_(n→+∞) ∫_0 ^1 x^2 sinx dx let integrate by parts I =∫_0 ^1 x^2 sinxdx =[−x^2 cosx]_0 ^1 +∫_0 ^1 2x cosxdx =−cos(1) +2 { [xsinx]_0 ^1 −∫_0 ^1 sinxdx} = −cos(1) +2{ sin(1) +[cosx]_0 ^1 } =−cos(1) +2{ sin(1) +cos(1)−1} =2sin(1) +cos(1) −2 lim_(n→+∞) S_n = 2sin(1) +cos(1) −2 .$
$l e t p u t S_{n} = \frac{1}{n^{3}} \sum_{k = 1}^{n} k^{2} s i n (\frac{k π}{n})$ $S_{n} = \frac{1}{n} \sum_{k = 1}^{n} {(\frac{k}{n})}^{2} s i n (\frac{k π}{n}) \to_{n \to + \infty} \int_{0}^{1} x^{2} s i n x d x$ $l e t i n t e g r a t e b y p a r t s$ $I = \int_{0}^{1} x^{2} s i n x d x = {[- x^{2} c o s x]}_{0}^{1} + \int_{0}^{1} 2 x c o s x d x$ $= - c o s (1) + 2 {{[x s i n x]}_{0}^{1} - \int_{0}^{1} s i n x d x}$ $= - c o s (1) + 2 {s i n (1) + {[c o s x]}_{0}^{1}}$ $= - c o s (1) + 2 {s i n (1) + c o s (1) - 1}$ $= 2 s i n (1) + c o s (1) - 2$ ${l i m}_{n \to + \infty} S_{n} = 2 s i n (1) + c o s (1) - 2 .$
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