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JEE Main 20231 FebruaryMorning Shift
Exam
JEE Main
Year
2023
Date
1 February
Shift
Morning Shift
Subject
Mathematics
Chapter
Definite Integration

JEE Main 2023 · 1 February · Morning Shift · Mathematics · Definite Integration

_ n → ∞ ((1)/(1 + n) + (1)/(2 + n) + (1)/(3 + n) + + (1)/(2n)) is...

\lim_{n \to \infty}(11+n+12+n+13+n+…+12n)\left(\dfrac{1}{1 + n} + \dfrac{1}{2 + n} + \dfrac{1}{3 + n} + \ldots + \dfrac{1}{2n}\right) is equal to :-

Options

  1. A00
  2. Blog⁡e2\log_{e}2
  3. Clog⁡e(32)\log_{e}\left(\dfrac{3}{2}\right)
  4. Dlog⁡e(23)\log_{e}\left(\dfrac{2}{3}\right)

Correct answer

B. log⁡e2\log_{e}2

Step-by-step solution

We have,

lim⁡n→∞(11+n+12+n+13+n+…+12n)\lim_{n \to \infty}\left(\dfrac{1}{1 + n} + \dfrac{1}{2 + n} + \dfrac{1}{3 + n} + \ldots + \dfrac{1}{2n}\right)

=lim⁡n→∞(11+n+21+n+31+n+….+1n+n)= \lim_{n \to \infty}\left(\dfrac{1}{1 + n} + \dfrac{2}{1 + n} + \dfrac{3}{1 + n} + \ldots . + \dfrac{1}{n + n}\right)

=lim⁡n→∞∑r=1n1n+r= \lim_{n \to \infty}\mathop{∑}_{r = 1}^{n}\dfrac{1}{n + r}

=lim⁡n→∞∑r=1n1n(11+rn)= \lim_{n \to \infty}\mathop{∑}_{r = 1}^{n}\dfrac{1}{n}\left(\dfrac{1}{1 + \dfrac{r}{n}}\right)

=∫0111+xdx= {∫}_{0}^{1}\dfrac{1}{1 + x}dx

=[log⁡e(1+x)]01= {\left[\log_{e}\left(1 + x\right)\right]}_{0}^{1}

=log⁡e(2)= \log_{e}\left(2\right)

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