calculate lim_(n→+∞) S_n with S_n = (1/n^4 ) Σ_(k=1) ^n (k^3 /(√((1+((k/n))^2 )^3 )))


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Question Number 42805 by maxmathsup by imad last updated on 02/Sep/18

$c a l c u l a t e {l i m}_{n \to + \infty} S_{n} w i t h$ $S_{n} = \frac{1}{n^{4}} \sum_{k = 1}^{n} \frac{k^{3}}{\sqrt{{(1 + {(\frac{k}{n})}^{2})}^{3}}}$
Commented by maxmathsup by imad last updated on 03/Sep/18

$w e h a v e S_{n} = \frac{1}{n} \sum_{k = 1}^{n} \frac{{(\frac{k}{n})}^{3}}{\sqrt{{(1 + {(\frac{k}{n})}^{2})}^{3}}} \Rightarrow S_{n} i s a R i e m a n s u m a n d$ ${l i m}_{n \to + \infty} S_{n} = \int_{0}^{1} \frac{x^{3}}{\sqrt{{(1 + x^{2})}^{3}}} d x = \int_{0}^{1} \frac{x^{3}}{{(\sqrt{1 + x^{2}})}^{3}} d x c h a n g e m e n \sqrt{1 + x^{2}} = t$ $g i v e 1 + x^{2} = t^{2} \Rightarrow x d x = t d t \Rightarrow \int_{0}^{1} \frac{x^{3}}{(\sqrt{{1 + x^{2})}^{3}}} d x = \int_{1}^{\sqrt{2}} \frac{(t^{2} - 1) t d t}{t^{3}}$ $= \int_{1}^{\sqrt{2}} \frac{t^{3} - t}{t^{3}} d t = \int_{1}^{\sqrt{2}} (1 - \frac{1}{t^{2}}) = {[t + \frac{1}{t}]}_{1}^{\sqrt{2}} = \sqrt{2} + \frac{1}{\sqrt{2}} - 2 = \frac{3}{\sqrt{2}} - 2 \Rightarrow$ ${l i m}_{n \to + \infty} S_{n} = \frac{3 - 2 \sqrt{2}}{\sqrt{2}} .$
	Answered by tanmay.chaudhury50@gmail.com last updated on 03/Sep/18
	$lim_(n→∞) S_n =lim n→∞ Σ_(k=1) ^n (1/n)((((k/n))^3 )/(√({1+((k/n))^2 }^3 ))) ∫_0 ^1 (x^3 /(√({1+x^2 }^3 )))dx =∫_0 ^1 (x^3 /((1+x^2 )^(3/2) ))dx t^2 =1+x^2 tdt=xdx ∫_1 ^(√2) (((t^2 −1)tdt)/t^3 ) ∫_1 ^(√2) 1−(1/t^2 ) dt ∣t+(1/t)∣_1 ^(√2) ((√2) −1)+((1/(√2))−1) =((2−(√2) +1−(√2))/(√2))=((3−2(√2))/(√2))$
	$Double subscripts: use braces to clarify$ $\int_{0}^{1} \frac{x^{3}}{\sqrt{{1 + x^{2}}^{3}}} d x = \int_{0}^{1} \frac{x^{3}}{{(1 + x^{2})}^{\frac{3}{2}}} d x$ $t^{2} = 1 + x^{2} t d t = x d x$ $\int_{1}^{\sqrt{2}} \frac{(t^{2} - 1) t d t}{t^{3}}$ $\int_{1}^{\sqrt{2}} 1 - \frac{1}{t^{2}} d t$ $∣ t + \frac{1}{t} ∣_{1}^{\sqrt{2}}$ $(\sqrt{2} - 1) + (\frac{1}{\sqrt{2}} - 1)$ $= \frac{2 - \sqrt{2} + 1 - \sqrt{2}}{\sqrt{2}} = \frac{3 - 2 \sqrt{2}}{\sqrt{2}}$
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