JEE MainMathematics

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+4 marks−1 if incorrectSingle correctPrevious-year question

Limit of a Finite Sum via Definite Integral

Q: Why do the integration limits become 0 to 1?

Because the lower limit is lim as n → ∞ of (r_min)/n = lim as n → ∞ of 1/n = 0, and the upper limit is lim as n → ∞ of (r_max)/n = lim as n → ∞ of n/n = 1.

lim as n → ∞ of [1/(1+n) + 1/(2+n) + 1/(3+n) + … + 1/2n] is equal to

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Alog_e 2Blog_e(3/2)Clog_e(2/3)D0

Step-by-step solutionView

Correct answer

The limit equals $\int_{0}^{1} \frac{1}{1 + x} d x = \log_{e} 2$ .

Option analysis

Why each option works or fails

A · log_e 2

The student correctly converts the limit of the Riemann sum into a definite integral $\int_{0}^{1} \frac{d x}{1 + x} = [\ln (1 + x)]_{0}^{1} = \ln 2$ . Correctly expressed the sum in terms of $\frac{r}{n}$ , factored out $\frac{1}{n}$ , and evaluated the resulting definite integral.

B · log_e(3/2)

The student mistakenly sets the upper integration limit or substitution range incorrectly, evaluating $\ln (1 + x)$ from $x = 1 / 2$ to $1$ or similar, leading to $\ln (3 / 2)$ . Identify the running index $r$ from $1$ to $n$ , which gives limits of integration from $\lim \frac{1}{n} = 0$ to $\lim \frac{n}{n} = 1$ .

C · log_e(2/3)

The student inverts the argument of the logarithm, possibly evaluating $[\ln (1 + x)]$ as $\ln (1) - \ln (1 + 1) = - \ln 2$ or miscomputing $\ln (1 + 1 / 2) - \ln (2) = \ln (2 / 3)$ . Evaluate the antiderivative properly at the upper limit minus the lower limit: $[\ln (1 + x)]_{0}^{1} = \ln (2) - \ln (1) = \ln 2$ .

D · 0

The student takes the limit of each term individually as $n \to \infty$ , getting $\lim_{n \to \infty} \frac{1}{r + n} = 0$ , and concludes the sum of $n$ such terms is 0. Because the number of terms grows with $n$ , you cannot distribute the limit across the sum; formulate it as a Riemann sum instead.

Reviewed route

Solution

StepWorking

01Given

The given expression is $S = \lim_{n \to \infty} [\frac{1}{n + 1} + \frac{1}{n + 2} + \frac{1}{n + 3} + \dots + \frac{1}{n + n}]$ .

02Goal

Evaluate the limit as $n \to \infty$ using Riemann sum integration.

03Approach

Factor out $\frac{1}{n}$ to express the general term as a function of $\frac{r}{n}$ , and convert the Riemann sum $\lim_{n \to \infty} \frac{1}{n} \sum f (\frac{r}{n})$ into $\int_{0}^{1} f (x) d x$ .

04Execute

$S = \lim_{n \to \infty} \sum_{r = 1}^{n} \frac{1}{n + r} = \lim_{n \to \infty} \frac{1}{n} \sum_{r = 1}^{n} \frac{1}{1 + \frac{r}{n}} = \int_{0}^{1} \frac{1}{1 + x} d x = [\ln (1 + x)]_{0}^{1} = \ln 2 - \ln 1 = \ln 2$ .

✓Verify

For finite $n$ , $\sum_{r = 1}^{n} \frac{1}{n + r} = H_{2 n} - H_{n} \approx (\ln (2 n) + γ) - (\ln n + γ) = \ln 2$ . The result matches.

Hints that build this answer step by step

How can the given sum be rewritten in sigma notation with a factor of $\frac{1}{n}$ pulled out?
$\lim_{n \to \infty} \frac{1}{n} \sum_{r = 1}^{n} \frac{1}{1 + \frac{r}{n}}$
What definite integral represents $\lim_{n \to \infty} \frac{1}{n} \sum_{r = 1}^{n} \frac{1}{1 + \frac{r}{n}}$ ?
$\int_{0}^{1} \frac{1}{1 + x} d x$
What is the value of $\int_{0}^{1} \frac{1}{1 + x} d x$ ?
$\log_{e} 2$

✓ Source and academic review

Question type: Single correct
Exam relevance: JEE Main · Mathematics
Academic status: Answer verified against the official NTA key
Source: Previous-year question
Editorial review: 7 September 2026

Quick checks

Students also ask

Why do the integration limits become 0 to 1?

Because the lower limit is $\lim_{n \to \infty} \frac{r_{\min}}{n} = \lim_{n \to \infty} \frac{1}{n} = 0$ , and the upper limit is $\lim_{n \to \infty} \frac{r_{\max}}{n} = \lim_{n \to \infty} \frac{n}{n} = 1$ .

Answer

The limit equals $\int_{0}^{1} \frac{1}{1 + x} d x = \log_{e} 2$ .

Why each option works or fails

A · correct: log_e 2 - The student correctly converts the limit of the Riemann sum into a definite integral $\int_{0}^{1} \frac{d x}{1 + x} = [\ln (1 + x)]_{0}^{1} = \ln 2$ . Correctly expressed the sum in terms of $\frac{r}{n}$ , factored out $\frac{1}{n}$ , and evaluated the resulting definite integral.
B: log_e(3/2) - The student mistakenly sets the upper integration limit or substitution range incorrectly, evaluating $\ln (1 + x)$ from $x = 1 / 2$ to $1$ or similar, leading to $\ln (3 / 2)$ . Identify the running index $r$ from $1$ to $n$ , which gives limits of integration from $\lim \frac{1}{n} = 0$ to $\lim \frac{n}{n} = 1$ .
C: log_e(2/3) - The student inverts the argument of the logarithm, possibly evaluating $[\ln (1 + x)]$ as $\ln (1) - \ln (1 + 1) = - \ln 2$ or miscomputing $\ln (1 + 1 / 2) - \ln (2) = \ln (2 / 3)$ . Evaluate the antiderivative properly at the upper limit minus the lower limit: $[\ln (1 + x)]_{0}^{1} = \ln (2) - \ln (1) = \ln 2$ .
D: 0 - The student takes the limit of each term individually as $n \to \infty$ , getting $\lim_{n \to \infty} \frac{1}{r + n} = 0$ , and concludes the sum of $n$ such terms is 0. Because the number of terms grows with $n$ , you cannot distribute the limit across the sum; formulate it as a Riemann sum instead.

Step-by-step solution

Given: The given expression is $S = \lim_{n \to \infty} [\frac{1}{n + 1} + \frac{1}{n + 2} + \frac{1}{n + 3} + \dots + \frac{1}{n + n}]$ .
Goal: Evaluate the limit as $n \to \infty$ using Riemann sum integration.
Approach: Factor out $\frac{1}{n}$ to express the general term as a function of $\frac{r}{n}$ , and convert the Riemann sum $\lim_{n \to \infty} \frac{1}{n} \sum f (\frac{r}{n})$ into $\int_{0}^{1} f (x) d x$ .
Execute: $S = \lim_{n \to \infty} \sum_{r = 1}^{n} \frac{1}{n + r} = \lim_{n \to \infty} \frac{1}{n} \sum_{r = 1}^{n} \frac{1}{1 + \frac{r}{n}} = \int_{0}^{1} \frac{1}{1 + x} d x = [\ln (1 + x)]_{0}^{1} = \ln 2 - \ln 1 = \ln 2$ .
Verify: For finite $n$ , $\sum_{r = 1}^{n} \frac{1}{n + r} = H_{2 n} - H_{n} \approx (\ln (2 n) + γ) - (\ln n + γ) = \ln 2$ . The result matches.

Limit of a Finite Sum via Definite Integral

What feels right?

The limit equals $\int_{0}^{1} \frac{1}{1 + x} d x = \log_{e} 2$ .

Why each option works or fails

Solution

01Given

02Goal

03Approach

04Execute

✓Verify

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