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JEE Advanced 2022 Paper 1 · Mathematics · Limits · medium

Evaluate lim n→∞prodk=1^n(1+(k^2/(n^2)^1/n.

Numerical answer

Evaluate lim⁡n→∞∏k=1n(1+k2n2)1/n\displaystyle\lim_{n\to\infty}\prod_{k=1}^{n}\left(1+\frac{k^2}{n^2}\right)^{1/n}. Give the answer correct to two decimal places.
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Answer: 1.30

Take logarithms: ln⁡L=lim⁡1n∑k=1nln⁡ ⁣(1+k2n2)=∫01ln⁡(1+x2) dx\ln L=\lim\frac1n\sum_{k=1}^n\ln\!\left(1+\frac{k^2}{n^2}\right)=\int_0^1\ln(1+x^2)\,dx. By parts, ∫01ln⁡(1+x2)dx=[xln⁡(1+x2)]01−∫012x21+x2dx=ln⁡2−2(1−π4)=ln⁡2−2+π2\int_0^1\ln(1+x^2)dx=\left[x\ln(1+x^2)\right]_0^1-\int_0^1\frac{2x^2}{1+x^2}dx=\ln2-2\left(1-\frac\pi4\right)=\ln2-2+\frac\pi2. Hence L=2eπ/2−2≈1.3021L=2e^{\pi/2-2}\approx1.3021, i.e. 1.301.30.
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