Question

$\lim_{n \to \infty} \frac{1}{ \sqrt{n} } ( \frac{1}{1+ \sqrt{3} } +\frac{1}{ \sqrt{3} + \sqrt{5} }+....+\frac{1}{ \sqrt{2n-1} + \sqrt{2n+1} })=?$

Answer 1

[tex]\frac{1}{\sqrt{n}}(\frac{1}{1+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{5}}+....+\frac{1}{\sqrt{2n-1}+\sqrt{2n+1}})=\\ \frac{1}{\sqrt{n}}(\frac{\sqrt{3}-1}{2}+\frac{\sqrt{5}-\sqrt{3}}{2}+....+\frac{\sqrt{2n+1}-\sqrt{2n-1}}{2})=\\ \frac{1}{\sqrt{n}}(\frac{-1+\sqrt{2n+1}}{2})=\\ \frac{\sqrt{n}}{n}(\frac{\sqrt{2n+1}-1}{2})=\\ \frac{\sqrt{2n^2+n}-\sqrt{n}}{2n} [/tex]