Question

lim n tend to infinity [1/√2n-1 + 1/√4n-2² +1/√6n-3²+ .......+ 1/n] = π/2

Answer 1

The question asks to prove if $Lim_{n \to \infin}[\frac{1}{\sqrt{2n-1^2}}+\frac{1}{\sqrt{4n-2^2}}+\frac{1}{\sqrt{6n-3^2}}+...+\frac{1}{n}]=\frac{\pi}{2}$

$Lim_{n \to \infin} \sum _{n=1} ^n\frac{1}{\sqrt{2n-r^2}}$

$Lim_{n \to \infin} \sum _{n=1} ^n\frac{1}{n\sqrt{\frac{2r}{n}- (\frac{r}{n})^2}}$

$\frac{r}{n}=x$

$\int_0^1 \frac{1}{\sqrt{2x-x^2}}dx=\int_0^1 \frac{1}{\sqrt{1-(x-1)^2}}dx$

Let x-1 =t, dx=dt

$\int_{-1}^0 \frac{dt}{\sqrt{1-t^2}}=[sin^{-1} t]_{-1}^0=0-[- \frac{\pi}{2}]= \frac{\pi}{2}$