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AP EAMCET Engineering 23 Apr 2019 Shift 1 Paper

Section: Mathematics

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Question : 76 of 160

Marks: +1, -0

\lim\limits_{n\to\infty}\frac{1}{n}

[(n+1)(n+2)\dots(2n)]^{1/n}

=

1
0
$\frac{2}{e}$
$\frac{4}{e}$

Solution:

Consider the integral,

P=\lim\limits_{n\to\infty}\frac{1}{n}

[(n+1)(n+2)\dots(2n)]^{1/n}

Apply log on both sides,

\log P=\lim\limits_{n\to\infty}\frac{1}{n}[\log \left(\frac{n+1}{n}\right)\left(\frac{n+2}{n}\right)

\dots\left(\frac{n+n}{n}\right)]

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

=\int\limits_{0}^{1} \log(1+x)

=\left[x \log(1+x)\right]_{0}^{1}-\int\limits_{0}^{1} \frac{x}{1+x}\,dx

Further simplify the above,

\log P = \log 2 - \left( x - \log(1+x) \right)_{0}^{1}

=\log 2 - 1 + \log 2

=\log 4 - 1

=\log \left( \frac{4}{e} \right)

This implies,

P = \frac{4}{e}

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