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JEE Main 25 Feb 2021 Shift 2 Solved Paper

Section: Mathematics

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Question : 74 of 90

Marks: +1, -0

\lim\limits_{n \to \infty} \left[ \frac{1}{n} + \frac{n}{(n+1)^2} + \frac{n}{(n+2)^2} + \dots + \frac{n}{(2n-1)^2} \right] \; \text{ is }

[25 Feb 2021 Shift 2]

1
$\frac{1}{2}$
$\;{1}/{3}$
$\;{1}/{4}$

Solution:

(b)

\; \text{ Let } \; L \;\; = \lim\limits_{n \to \infty} \left[ \frac{1}{n} + \frac{n}{(n+1)^2} + \frac{n}{(n+2)^2} + \dots \right]

\; \text{ or } \; L \;\; = \lim\limits_{n \to \infty} \left[ \frac{n}{(2n-1)^2} \right] \frac{n}{(n+0)^2} + \frac{n}{(n+1)^2} + \dots +

\;\; = \lim\limits_{n \to \infty} \left[ \frac{n}{(n+n-1)^2} + \frac{n}{(n+n)^2} - \frac{n}{(n+n)^2} \right]

\;\; - \lim\limits_{n \to \infty} \left[ \frac{n}{(n+n)^2} \right]

\; = \lim\limits_{n \to \infty} \sum\limits_{r=0}^{n} \frac{n}{(n+1)^2} + \dots + \frac{n}{(n+n)^2} ]

\;\; = \lim\limits_{n \to \infty} \sum\limits_{r=0}^{n} \frac{n}{(n+r)^2} - 0 \quad \left(\text{since } \lim\limits_{n \to \infty} \frac{1}{n} = 0\right)

Now, for solving limit summation, we integrate it using some replacement.

L = \lim\limits_{n \to \infty} \sum\limits_{r=0}^{n} \frac{1}{r\left(1+\frac{r}{n}\right)}

Take

\frac{r}{n}

as

x

and

\frac{1}{n}

as

dx

.

Lower limit is obtained by putting

r=0

in

\frac{r}{n}

, we get Lower limit

=0

Upper limit is obtained by putting

r=n

in

\frac{r}{n}

, we get

Upper limit = 1

\therefore \; L = \int\limits_{0}^{1} \frac{1}{(1+x)^2} \, dx = \left[ \frac{-1}{1+x} \right]_{0}^{1} = -\left(\frac{1}{2} - 1\right) = \frac{1}{2}

\therefore \; L = \frac{1}{2}

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