QJEEAdv18Maf331: JEE Advanced 2018 Mathematics - Class 12 JEEAdv18Maf331 Solution: For each positive integer n, let . For x R, let [x] be the…

Question

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For each positive integer n, let

${y_n} = {1 \over n}(n + 1)(n + 2)...{(n + n)^{{1 \over n}}}$ .

For x

\in

R, let [x] be the greatest integer less than or equal to x. If

\mathop {\lim }\limits_{n \to \infty } {y_n} = L

, then the value of [L] is ..............

Accepted Answer

We have,

{y_n} = {1 \over n}(n + 1)(n + 2)...{(n + n)^{1/n}}

and

\mathop {\lim }\limits_{n \to \infty } {y_n} = L

\Rightarrow L = \mathop {\lim }\limits_{n \to \infty } {1 \over n}{[(n + 1)(n + 2)(n + 3)...(n + n)]^{1/n}}

\Rightarrow L = \mathop {\lim }\limits_{n \to \infty } {\left[ {\left( {1 + {1 \over n}} \right)\left( {1 + {2 \over n}} \right)\left( {1 + {3 \over n}} \right)...\left( {1 + {n \over n}} \right)} \right]^{{1 \over n}}}

\Rightarrow \log L = \mathop {\lim }\limits_{n \to \infty } {1 \over n}\left[ {\log \left( {1 + {1 \over n}} \right) + \log \left( {1 + {2 \over n}} \right)...\log \left( {1 + {n \over n}} \right)} \right]

\Rightarrow \log L = \mathop {\lim }\limits_{n \to \infty } {1 \over n}\sum\limits_{r = 1}^n {\log \left( {1 + {r \over n}} \right)}

\Rightarrow \log L = \int_0^1 {_{II}^1 \times \mathop {\log }\limits_I } (1 + x)\,dx

\Rightarrow \log L = (x.\log (1 + x)_0^1 - \int_0^1 {\left[ {{d \over {dx}}(\log (1 + x)\int {dx} } \right]} dx

[by using integration by parts]

\Rightarrow \log L = [x\log (1 + x)]_0^1 - \int_0^1 {{x \over {1 + x}}} dx

\Rightarrow \log L = \log 2 - \int_0^1 {\left( {{{x + 1} \over {x + 1}} - {1 \over {x + 1}}} \right)} dx

\Rightarrow \log L = \log 2 - [x]_0^1 + [\log (x + 1)]_0^1

\Rightarrow \log L = \log 2 - 1 + \log 2 - 0

\Rightarrow \log L = \log 4 - \log e = \log {4 \over e}

\Rightarrow L = {4 \over e}

\Rightarrow [L] = \left[ {{4 \over e}} \right] = 1

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