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JEE Mathematics 2023 Question with Solution

If lim⁡n→∞(11+n+12+n+13+n+⋯+12n)\lim_{n \to \infty} \left( \frac{1}{1+n} + \frac{1}{2+n} + \frac{1}{3+n} + \dots + \frac{1}{2n} \right) is equal to:

  • A

    log⁡e2\log_e 2

  • B

    log⁡e(32)\log_e \left( \frac{3}{2} \right)

  • C

    log⁡e(23)\log_e \left( \frac{2}{3} \right)

  • D

    Option unavailable in scraped input

Answer

Correct answer:A

Step-by-step solution

Standard Method

Given:

lim⁡n→∞(11+n+12+n+⋯+1n+n)\lim_{n\to\infty} \left(\frac{1}{1+n}+\frac{1}{2+n}+\cdots+\frac{1}{n+n}\right)

Find: The value of the limit.

Rewrite the sum as

lim⁡n→∞∑r=1n1n+r\lim_{n\to\infty} \sum_{r=1}^{n} \frac{1}{n+r}

Then factor out 1n\frac{1}{n}:

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

This is the Riemann sum for

∫0111+x dx\int_0^1 \frac{1}{1+x} \, dx

Therefore,

∫0111+x dx=[ln⁡(1+x)]01=ln⁡2\int_0^1 \frac{1}{1+x} \, dx = [\ln(1+x)]_0^1 = \ln 2

Therefore, the value of the limit is log⁡e2\log_e 2. The correct option is A.

Riemann Sum Interpretation

The expression

∑r=1n1n(11+rn)\sum_{r=1}^{n} \frac{1}{n}\left(\frac{1}{1+\frac{r}{n}}\right)

matches the standard form of a Riemann sum

∑r=1nf(rn)1n\sum_{r=1}^{n} f\left(\frac{r}{n}\right)\frac{1}{n}

with

f(x)=11+xf(x)=\frac{1}{1+x}

Hence, as n→∞n \to \infty,

∑r=1n1n(11+rn)→∫0111+x dx\sum_{r=1}^{n} \frac{1}{n}\left(\frac{1}{1+\frac{r}{n}}\right) \to \int_0^1 \frac{1}{1+x} \, dx

Now evaluate the integral:

∫0111+x dx=[ln⁡(1+x)]01=ln⁡2−ln⁡1=ln⁡2\int_0^1 \frac{1}{1+x} \, dx = \left[\ln(1+x)\right]_0^1 = \ln 2 - \ln 1 = \ln 2

So the required limit equals log⁡e2\log_e 2. Note that the listed options list is incomplete, but the solution explicitly marks A as correct and evaluates the limit as log⁡e2\log_e 2.

Common mistakes

  • Treating the sum as a harmonic series tail without converting it into a Riemann sum. This misses the factor 1n\frac{1}{n} structure. Rewrite each term as 1n⋅11+r/n\frac{1}{n}\cdot\frac{1}{1+r/n} first.

  • Using wrong limits of integration. Since rn\frac{r}{n} runs from values near 00 to 11, the integral is from 00 to 11, not from 11 to nn.

  • Forgetting that ln⁡1=0\ln 1 = 0 and making an error while evaluating [ln⁡(1+x)]01\left[\ln(1+x)\right]_0^1. Substitute both limits carefully to get ln⁡2−ln⁡1=ln⁡2\ln 2 - \ln 1 = \ln 2.

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