Question

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

Answer 1

$\underset{n\to \infty }{lim}\frac{2n^2+n+1}{(n+1)+(n+2)+(n+3)+....+2n}$

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

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

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

$\Rightarrow \underset{n\to \infty }{lim}\frac{2(2n^2+n+1)}{2n^2+n^2+n}=\underset{n\to \infty }{lim}\frac{2(2n^2+n+1)}{3n^2+n}$

$=\underset{n\to \infty }{lim}\frac{2(2n^2(2+\frac{1}{n}+\frac{1}{n^2})}{n^2(3+\frac{1}{n})}$

$=\frac{2(2+0+0}{3}=\frac{4}{3}$.