Question

$\displaystyle \lim_{n \to \infty}$$\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

The Correct Option is B

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