Question

What is the limiting value of the expression
$\frac{\{(n+1)(n+2) \ldots \ldots \ldots(2 n)\}^{\frac{1}{n}}}{n}$ when $n$ tends to infinity?

• A. $\frac{1}{2}$
• B. e
• C. $\frac{2}{ e }$
• D. $\frac{4}{ e }$

Answer 1

Answer: (D)

Let $P =\frac{\{( n +1)( n +2) \ldots \ldots .(2 n )\}^{\frac{1}{ n }}}{ n }$
Then $\log P =\frac{1}{ n }[\log ( n +1)+\log ( n +2)+\ldots \ldots+\log 2 n ]-\log n$
$=\frac{1}{ n }[\{\log ( n +1)-\log n \}+\{\log ( n +2)-\log n \} +\ldots+\{\log 2 n -\log n \}]$
$=\frac{1}{ n }\left[\log \left(\frac{ n +1}{ n }\right)+\log \left(\frac{ n +2}{ n }\right)+\ldots . .+\log \left(\frac{ n + n }{ n }\right)\right]$
$=\frac{1}{ n }\left[\log \left(1+\frac{1}{ n }\right)+\log \left(1+\frac{2}{ n }\right)+\ldots \ldots .+\log \left(1+\frac{ n }{ n }\right)\right]$
$=\frac{1}{ n } \displaystyle\sum_{ r =1}^{ n } \log \left(1+\frac{ r }{ n }\right)=\int_\limits{0}^{1} \log (1+ x ) dx$
$=[ x \log (1+ x )]_{0}^{1}-\int_\limits{0}^{1} \frac{ x }{1+ x } dx$
[Integrating by parts taking 1 as second function]
$=[\log 2-0]-\int_\limits{0}^{1}\left[1-\frac{1}{1+ x }\right] dx$
$=\log 2-[x-\log (x+1)]_{0}^{1}=2 \log 2-1=\log 4-\log e$
$\therefore \log P =\log \frac{4}{ e } \Rightarrow P =\frac{4}{ e }$