JEE 2023 — Mathematics PYQ

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Question

JEE 2023 — Mathematics PYQ

JEE | Mathematics | 2023

\lim_{n \to \infty} \left[ \frac{1}{1+n} + \frac{1}{2+n} + \frac{1}{3+n} + \dots + \frac{1}{2n} \right]

Choose the correct answer:

Explanation

Solution

1. Generalize the Series

We can rewrite the sum using sigma notation:

L = \lim_{n \to \infty} \sum_{r=1}^{n} \frac{1}{r+n}

Divide the numerator and denominator by $n$ :

L = \lim_{n \to \infty} \sum_{r=1}^{n} \frac{1}{n} \left( \frac{1}{\frac{r}{n} + 1} \right)

2. Convert to Definite Integral

Using the rule $\lim_{n \to \infty} \sum \frac{1}{n} f(\frac{r}{n}) = \int_{0}^{1} f(x) dx$ :

Replace $\frac{1}{n}$ with $dx$ .
Replace $\frac{r}{n}$ with $x$ .
Lower limit: $\frac{r_{min}}{n} = \frac{1}{n} \to 0$ .
Upper limit: $\frac{r_{max}}{n} = \frac{n}{n} \to 1$ .

L = \int_{0}^{1} \frac{1}{1+x} dx

3. Evaluate the Integral

L = [\log_e(1+x)]_0^1 = \log_e(1+1) - \log_e(1+0)

L = \log_e 2 - 0 = \log_e 2

Correct Option: (1)

Choose the correct answer:

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