Question

$\underset{n\to \infty }{lim}\frac{1}{\sqrt{4n^2-1}}+\frac{1}{\sqrt{4n^2-2^2}}+.....\frac{1}{\sqrt{3n^2}}$

Answer 1

$\underset{n\to \infty }{lim}\frac{1}{\sqrt{4n^2-1}}+\frac{1}{\sqrt{4n^2-2^2}}+...+\frac{1}{\sqrt{4n^2-n^2}}$

$=\underset{n\to \infty }{lim}\frac{1}{n\sqrt{4-1/n^2}}+\frac{1}{n\sqrt{4-4/n^2}}+...+\frac{1}{n\sqrt{3}}$

as $n\to \infty \frac{1}{n}\to 0;\frac{1}{n^2}\to 0$

$=\frac{0}{\sqrt{4-0}}+\frac{0}{\sqrt{4-0}}+....+\frac{0}{\sqrt{3}}$

$=0$