Question

Solve

$\underset{n\to \infty }{lim}\left\{\frac{1}{n}+\frac{1}{\sqrt{n^2-1^2}}+\frac{1}{\sqrt{n^2-2^2}}+...+\frac{1}{\sqrt{n^2-{\left(n-1\right)}^2}}\right\}$

Answer 1

given $S_n=\frac{1}{\sqrt{(n{)}^2-0^2}}+\frac{1}{\sqrt{(n{)}^2-1^2}}+\frac{1}{\sqrt{(n{)}^2}-2^2}+..........+\frac{1}{\sqrt{(n{)}^2}-(n-1{)}^2}$

$=\sum _{r=0}^{n-1}\frac{1}{\sqrt{(n{)}^2-r^2}}=\sum _{r=0}^{n-1}\frac{1}{n}\frac{1}{\sqrt{1-(r/n{)}^2}}$

Now $\underset{n\to \infty }{lim}{S}_n={\int }_0^1\frac{dx}{\sqrt{1-x^2}}=si{n}^{-1}x{|}_0^1=\frac{\pi }{2}$