let-S-n-k-1-n-1-k-2-n-2-calculate-lim-n-S-n-

Question Number 43546 by maxmathsup by imad last updated on 11/Sep/18

l e t S_{n} = \sum_{k = 1}^{n} \frac{1}{\sqrt{k^{2} + n^{2}}} c a l c u l a t e {l i m}_{n \to + \infty} S_{n}

Answered by behi83417@gmail.com last updated on 12/Sep/18

S_{n} = \sum_{1}^{n} \frac{1}{n} . \frac{1}{\sqrt{1 + {(\frac{k}{n})}^{2}}} = \int_{0}^{1} \frac{d x}{\sqrt{1 + x^{2}}} =

= l n {(x + \sqrt{1 + x^{2}})}_{0}^{1} = l n (\sqrt{2} + 1) .

Commented by maxmathsup by imad last updated on 12/Sep/18

c h a 7 g e m e n t x = t a n θ g i v e \int_{0}^{1} \frac{d x}{\sqrt{1 + x^{2}}} = \int_{0}^{\frac{π}{4}} \frac{1 + {t a n}^{2} θ}{\sqrt{1 + {t a n}^{2} θ}} d θ

= \int_{0}^{\frac{π}{4}} \sqrt{1 + {t a n}^{2} θ} = \int_{0}^{\frac{π}{4}} \frac{d θ}{c o s θ} =_{t a n (\frac{θ}{2}) = u} \int_{0}^{\sqrt{2} - 1} \frac{1}{\frac{1 - t^{2}}{1 + t^{2}}} \frac{2 d u}{1 + u^{2}}

= \int_{0}^{\sqrt{2} - 1} {\frac{1}{1 - u} + \frac{1}{1 + u}} d u = {[l n ∣ \frac{1 + u}{1 - u} ∣]}_{0}^{\sqrt{2} - 1} = l n ∣ \frac{\sqrt{2}}{2 - \sqrt{2}} ∣

= l n (\frac{1}{\sqrt{2} - 1}) = l n (1 + \sqrt{2}) \Rightarrow {l i m}_{n \to + \infty} S_{n} = l n (1 + \sqrt{2})

a n o t h e r w a y \int_{0}^{1} \frac{d x}{\sqrt{1 + x^{2}}} = {[a r g s h (x)]}_{0}^{1} = {[l n (x + \sqrt{1 + x^{2}})]}_{0}^{1} = l n (1 + \sqrt{2}) .