If ₙ → ∞} (n+r)- n/n=2 ≤ft( 2-1/2 ), then ₙ → ∞} 1/n^λ ≤ft[ (n+1)^λ (n+2)^λ (n+n)^λ ]¹/n} is equal to

Correct Answer is : (4e)λ≤ft( 4/e )^λ(e4)λ

Correct Answer is : (4e)λ≤ft( 4/e )^λ(e4)λ

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CGPET 2014 Solved Paper

Section: Mathematics

Question : 130 of 150

Marks: +1, -0

\lim\limits_{n \to \infty} \sum \frac{\log (n+r)-\log n}{n}=2 \left( \log 2-\frac{1}{2} \right)

\lim\limits_{n \to \infty} \frac{1}{n^\lambda} \left[ (n+1)^\lambda (n+2)^\lambda \dots (n+n)^\lambda \right]^{1/n}

Solution:

Given,

\lim\limits_{n \to \infty} \sum \frac{\log (n+r)-\log n}{n}=2 \left( \log 2-\frac{1}{2} \right)

\lim\limits_{n \to \infty} \sum \frac{1}{n} \log \left( 1+\frac{r}{n} \right)=2 \left( \log 2-\frac{1}{2} \right)

Let

A=\lim\limits_{n \to \infty} \frac{1}{n^\lambda} \left[ (n+1)^\lambda (n+2)^\lambda \dots (n+n)^\lambda \right]^{1/n}

=\lim\limits_{n \to \infty} \left[ \left( 1+\frac{1}{n} \right)^\lambda \left( 1+\frac{2}{n} \right)^\lambda \dots \left( 1+\frac{n}{n} \right)^\lambda \right]^{1/n}

On taking log both sides, we get

\log A=\lim\limits_{n \to \infty} \frac{1}{n} \left[ \log \left( 1+\frac{1}{n} \right)^\lambda +\log \left( 1+\frac{2}{n} \right)^\lambda \dots \right.

+\dots +\log \left( 1+\frac{n}{n} \right)^\lambda

\Rightarrow \quad \log A=\lim\limits_{n \to \infty} \frac{1}{n} \sum\limits_{r=1}^{n} \lambda \log \left( 1+\frac{r}{n} \right)

\Rightarrow \quad \log A=\lambda \lim\limits_{n \to \infty} \sum\limits_{r=1}^{n} \frac{1}{n} \log \left( 1+\frac{r}{n} \right)

\Rightarrow \quad \log A=2 \lambda \left( \log 2-\frac{1}{2} \right)

\Rightarrow \quad \quad \left[ \text{ from Eq. (i) } \right)

\Rightarrow \quad \log A=\log 4^\lambda -\lambda

\Rightarrow \quad \log A=\log 4^\lambda -\lambda \log e

\therefore \quad \log A=\log \frac{4^\lambda}{e^\lambda}

\Rightarrow \quad A=\left( \frac{4}{e} \right)^\lambda

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