Question

Find the following limit.
$\underset{n\to \infty }{lim}\frac{n^2+n+1}{1+3+5+...+(2n-1)}.$

Answer 1

$\underset{n\to \infty }{lim}\frac{n^2+n+1}{1+3+5+....+(2n-1)}$

here $S_n=1+3+5+.....+(2n-1)$ is an A.P.

So $S_n=\frac{n}{2}[2a+(n-1\left)d\right]$

$=\frac{n}{2}[2+(n-1\left)2\right]$

$=\frac{n}{2}[2+2n-2]$

$=n^2$

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

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

$\Rightarrow 1$.