Question

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

• A. $1$
• B. $\frac{4}{3}$
• C. $\frac{3}{4}$
• D. $\infty$

Answer 1

The correct option is A $1$

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