Question

Evaluate: $\underset{n\to \mathrm{\infty }}{lim}\left[\frac{1}{n}+\frac{n}{(n+1{)}^2}+\frac{n}{(n+2{)}^2}+......+\frac{n}{(2n-1{)}^2}\right]$

• A. $1$
• B. $\frac{1}{3}$
• C. $\frac{1}{2}$
• D. $\frac{1}{4}$

Answer 1

The correct option is D

$\frac{1}{4}$

Explanation for the correct option:

Given: $\underset{n\to \mathrm{\infty }}{lim}\left[\frac{1}{n}+\frac{n}{(n+1{)}^2}+\frac{n}{(n+2{)}^2}+......+\frac{n}{(2n-1{)}^2}\right]$

Simplifying and solving above limit, we get

$$\begin{gathered} =\underset{n\to \mathrm{\infty }}{lim}\left[\frac{1}{{\left(n+0\right)}^2}+\frac{n}{(n+1{)}^2}+\frac{n}{(n+2{)}^2}+......+\frac{n}{(2n-1{)}^2}\right] \\ =\underset{n\to \mathrm{\infty }}{lim}\sum _{r=0}^{n-1}\left[\frac{n}{{\left(n+r\right)}^2}\right] \\ =\underset{n\to \mathrm{\infty }}{lim}\sum _{r=0}^{n-1}\left[\frac{n}{n^2{\left(1+{\displaystyle \frac{r}{n}}\right)}^2}\right] \\ =\underset{n\to \mathrm{\infty }}{lim}\underset{r=0}{\overset{n-1}{\sum \frac{1}{n}}}\left[\frac{1}{{}^{}{\left(1+{\displaystyle \frac{r}{n}}\right)}^2}\right] \end{gathered}$$

Replace $\frac{r}{n}=x\&\frac{1}{n}=dx$

When $$\begin{gathered} r=0,x=0 \\ r=n,x=1 \end{gathered}$$

$$\begin{gathered} \Rightarrow {\int }_0^1\frac{dx}{{\left(1+x\right)}^2} \\ \Rightarrow {\int }_0^1{\left(1+x\right)}^{-2}dx \end{gathered}$$

We know that, $\int x^{n}dx=\frac{x^{n+1}}{n+1}+c$

$$\begin{gathered} =\left[\frac{{\left(1+x\right)}^{-2}}{-2}\right]\begin{array}{l}1\\ 0\end{array} \\ =-\frac{1}{2}\left[\frac{1}{2}-1\right]\begin{array}{l}1\\ 0\end{array} \\ =-\frac{1}{2}\left(-\frac{1}{2}\right) \\ =\frac{1}{4} \end{gathered}$$

Therefore, the value of the limit is $\frac{1}{4}$

Hence, option (D) is the correct answer.