calculate-lim-n-S-n-with-S-n-1-n-4-k-1-n-k-3-1-k-n-2-3-

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

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}}