JEE Advanced 2018 Paper 1Question 1
For each positive integer $n$, let $$y_n=\frac{1}{n}\left((n+1)(n+2)\cdots(n+n)\right)^{\frac{1}{n}}.$$ For $x\in\mathbb{R}$, let $[x]$ be the greatest integer less than or equal to $x$. If $\lim_{n\to\infty}y_n=L$, then the value of $[L]$ is _____ .
Limits & Continuity
Answer$1$
Solution
Concept. Take logs: the product becomes a sum, which is a Riemann sum. Formulas used. (1) $\lim_{n\to\infty}\frac1n\sum_{r=1}^{n}g\big(\frac rn\big)=\int_0^1g(x)\,dx$(2) $\int\ln(1+x)\,dx=(1+x)\ln(1+x)-(1+x)$Step 1 - Rewrite. $$y_n=\Big(\prod_{r=1}^{n}\frac{n+r}{n}\Big)^{1/n}=\Big(\prod_{r=1}^{n}\Big(1+\frac rn\Big)\Big)^{1/n}$$ Step 2 - Log. $\ln y_n=\frac1n\sum_{r=1}^{n}\ln\big(1+\frac rn\big)$. Step 3 - Use (1), (2). $$\ln L=\int_0^1\ln(1+x)\,dx=(2\ln2-2)-(0-1)=2\ln2-1$$ Step 4 - Value. $L=e^{2\ln2-1}=\frac4e\approx\frac{4}{2.718}\approx1.47$. Step 5 - Floor. $[L]=1$. Answer: 1