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JEE Advanced 2018 · Paper 1, Mathematics question 11

For each positive integer nn, let

yn=1n((n+1)(n+2)⋯(n+n))1n.y_n = \frac{1}{n}\left((n+1)(n+2)\cdots(n+n)\right)^{\frac{1}{n}}.

For x∈Rx \in \mathbb{R}, let [x][x] be the greatest integer less than or equal to xx. If lim⁡n→∞yn=L\displaystyle\lim_{n \to \infty} y_n = L, then the value of [L][L] is _____ .

Answer and worked solution

Official answer: 1

Take logs.
ln⁡yn=1n∑k=1nln⁡(n+k)−ln⁡n=1n∑k=1nln⁡(1+kn).\ln y_n=\frac1n\sum_{k=1}^n\ln(n+k)-\ln n=\frac1n\sum_{k=1}^n\ln\left(1+\frac kn\right).

Riemann sum.
ln⁡L=∫01ln⁡(1+x) dx=[(1+x)ln⁡(1+x)−x]01=2ln⁡2−1.\ln L=\int_0^1\ln(1+x)\,dx=\big[(1+x)\ln(1+x)-x\big]_0^1=2\ln2-1.

Value. L=4e≈1.47L=\dfrac4e\approx1.47, so [L]=1[L]=1.

Answer: 1

Topics:Limit of a sum
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