Question

Find the following limit.
$\underset{n\to \infty }{lim}(\frac{1}{n^2}+\frac{2}{n^2}+...+\frac{n-1}{n^2}).$

Answer 1

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

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

$=\underset{n\to \infty }{lim}\frac{(n-1)n}{2n^2}[\because 1+2+3+.....+n=\frac{n(n+1)}{2}]$

$\Rightarrow {lim}_{n\to \infty }\frac{n(1-\frac{1}{n})\cdot n}{2n^2}$

$\Rightarrow \frac{1-\frac{1}{\infty }}{2}=\frac{1}{2}$.