Question

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

• A. $1$
• B. $4/3$
• C. $3/4$
• D. $\infty$

Answer 1

The correct option is A $1$

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

Denominator can be written as $1+3+5+...+(2n-1)=\sum (2n-1)$

$\sum 2n-\sum 1\Rightarrow 2\sum n-n$

$\Rightarrow 2\frac{n(n+1)}{2}-n$

$\Rightarrow n^2+n-n=n^2$

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

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

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

$=1+\frac{1}{\infty }+\frac{1}{{\infty }^2}$

$=1$