Answer to Question #203180 in Real Analysis for Rajkumar

Question

Prove that

lim n→∞ [ 1/ √(2n-1) + 1/ √(4n-2²) + 1/ √(6n-3²) +.... + 1/n ] = π /2

Answer 1

$lim _{n→∞} [ \frac{1}{\sqrt{2n-1}}+\frac{1}{\sqrt{4n-2^2}}+\frac{1}{\sqrt{6n-3^2}}+•••+\frac{1}{n}]=lim _{n→∞}\sum_{k=1 } ^n [ \frac{1}{\sqrt{2kn-k^2}}]\\ =lim _{n→∞}\sum_{k=1 } ^n \frac{1}{n} \frac{1}{\sqrt{2(\frac{k}{n})-(\frac{k}{n}) ^2}}\\ \text{By integral test,}\\ \int_0^1 \frac{1}{\sqrt{2x-x^2}}dx\\ =\int_0^1 \frac{1}{\sqrt{1-(x-1)^2}}dx\\ Put \space x-1=t\implies dx=dt\\ \int_{-1}^0 \frac{1}{\sqrt{1-t^2}}dt\\ =sin^{-1}t]_{-1}^0\\ =sin^{-1}(0)-sin^{-1}(-1)\\ =0-(-\frac{\pi}{2})\\ =\frac{\pi}{2}$