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JEE MAIN 2021
25-02-2021 S2
Question
$\lim _{n \rightarrow \infty}\left[\frac{1}{n}+\frac{n}{(n+1)^2}+\frac{n}{(n+2)^2}+\ldots . .+\frac{n}{(2 n-1)^2}\right]$ is equal to
Select the correct option:
A
$\frac{1}{2}$
B
1
C
$\frac{1}{3}$
D
$\frac{1}{4}$
✓ Correct! Well done.
✗ Incorrect. Try again or view the solution.
Solution
$\begin{aligned} & \lim _{n \rightarrow \infty}\left[\frac{1}{n}+\frac{n}{(n+1)^2}+\frac{n}{(n+2)^2}+\ldots . .+\frac{n}{(2 n-1)^2}\right] \\ & =\lim _{n \rightarrow \infty} \sum_{r=0}^{n-1} \frac{n}{(n+r)^2}=\lim _{n \rightarrow \infty} \sum_{r=0}^{n-1} \frac{n}{n^2+2 n r+r^2} \\ & =\lim _{n \rightarrow \infty} \frac{1}{n} \sum_{r=0}^{n-1} \frac{1}{(r / n)^2+2(r / n)+1} \\ & =\int_0^1 \frac{d x}{(x+1)^2}=\left[\frac{-1}{(x+1)}\right]_0^1=\frac{1}{2}\end{aligned}$
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