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Question Number 30757 by abdo imad last updated on 25/Feb/18

f i n d {l i m}_{n \to \infty} (\frac{1}{n} + \frac{1}{\sqrt{n^{2} - 1}} + \dots . + \frac{1}{\sqrt{n^{2} - {(n - 1)}^{2}}})

Commented by abdo imad last updated on 28/Feb/18

l e t p u t w_{n} = \frac{1}{n} + \frac{1}{\sqrt{n^{2} - 1}} + \dots + \frac{1}{\sqrt{n^{2} - {(n - 1)}^{2}}}

w_{n} = \frac{1}{n} (1 + \frac{1}{\sqrt{1 - \frac{1^{2}}{n^{2}}}} + \frac{1}{\sqrt{1 - \frac{2^{2}}{n^{2}}}} + \dots . . + \frac{1}{\sqrt{1 - \frac{{(n - 1)}^{2}}{n^{2}}}})

= \frac{1}{n} \sum_{k = 0}^{n - 1} \frac{1}{\sqrt{1 - {(\frac{k}{n})}^{2}}} s o w_{n} i s a R i e m a n s u m a n d

{l i m}_{n \to \infty} w_{n} = \int_{0}^{1} \frac{d x}{\sqrt{1 - x^{2}}} = {[a r c s i n x]}_{0}^{1} = \frac{π}{2} .