For each positive integer \(n\), let \(y_{n}=\frac{1}{n}((n+1)(n+2) \cdots(n+n))^{\frac{1}{n}}\). For \(x \in \mathbb{R}\), let \([x]\) be the greatest integer less than or equal to \(x\). If \(\lim _{n \rightarrow \infty} y_{n}=L\), then the value of \([L]\) is ____.

JEE Advanced · 2018 · 20 May 2018 · Paper 1 · Limits XII

For each positive integer

n

, let

y_{n} = \frac{1}{n} ((n + 1) (n + 2) \dots (n + n))^{\frac{1}{n}}

. For

x \in 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 ____.

Answer: 1

y_{n} = \frac{1}{n} [(n + 1) (n + 2) \dots (2 n)]^{1/ n} = [\prod_{k = 1}^{n} (1 + \frac{k}{n})]^{1/ n}

.

Taking logarithm:

ln y_{n} = \frac{1}{n} \sum_{k = 1}^{n} ln (1 + \frac{k}{n})

.

As

n \to \infty

, this Riemann sum converges to

\int_{0}^{1} ln (1 + x) d x

\int_{0}^{1} ln (1 + x) d x = [x ln (1 + x)]_{0}^{1} - \int_{0}^{1} \frac{x}{1 + x} d x = ln 2 - \int_{0}^{1} (1 - \frac{1}{1 + x}) d x

= ln 2 - [x - ln (1 + x)]_{0}^{1} = ln 2 - (1 - ln 2) = 2 ln 2 - 1 = ln 4 - 1

.

So

L = e^{l n 4 - 1} = \frac{4}{e} \approx 1.47

.

Therefore

[L] = 1

JEE Advanced · 2018 · 20 May 2018 · Paper 1 · Numerical answer