Answer to Question #285787 in Real Analysis for Reens

Question

prove that lim n->infinity [1/√2n-1 +1/√4n-2^2 +1/√6n-3^2 +.......+1/n]=π/2

Answer 1

$\lim_{n\to \infty}\left[ \frac{1}{\sqrt{2n-1}}+\frac{1}{\sqrt{4n-2^2}}+\frac{1}{\sqrt{6n-3^2}}+\cdots +\frac{1}{n}\right]\\ \lim_{n\to \infty} \sum_{r=1}^n \frac{1}{\sqrt{2rn-r^2}}\\ \lim_{n\to \infty} \sum_{r=1}^n \frac{1}{n\sqrt{\frac{2r}{n}-(\frac{r}{n})^2}}\\$

Let $x=\frac{r}{n}, dx=\frac{1}{n}$

The limit becomes:

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

Let $x-1=t, dx=dt$ . When $x=0, t=-1, x=1, t=0.$

$\int_{-1}^0\frac{dt}{\sqrt{1-t^2}}=\sin^{-1}t \Biggr |_{-1}^0=\sin^{-1}0-\sin^{-1}(-1)=0-_-\frac{\pi}{2}=\frac{\pi}{2}$

Hence,

$\lim_{n\to \infty}\left[ \frac{1}{\sqrt{2n-1}}+\frac{1}{\sqrt{4n-2^2}}+\frac{1}{\sqrt{6n-3^2}}+\cdots +\frac{1}{n}\right]=\frac{\pi}{2}$