Question

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

• A. $\frac{1}{2}$
• B. $1$
• C. $-\frac{1}{2}$
• D. $-1$

Answer 1

Answer: (B)

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