Question

Find the value of the following limit.
\[\underset{n\to \infty }{\mathop{\lim }}\,{{\left( \dfrac{\left( n+1 \right)\left( n+2 \right).....\left( 3n \right)}{{{n}^{2n}}} \right)}^{\dfrac{1}{n}}}\]
\[\left( a \right)\dfrac{18}{{{e}^{4}}}\]
\[\left( b \right)\dfrac{27}{{{e}^{2}}}\]
\[\left( c \right)\dfrac{9}{{{e}^{2}}}\]
\[\left( d \right)3\log 3-2\]

Answer 1

Hint:We start by letting our limit as y. We have \[\dfrac{1}{n}\] in the power. So, we apply log on both sides to simplify this. We will use \[\log \left( ab \right)=\log a+\log b\] to convert this limit into summation.
\[\ln y=\underset{n\to \infty }{\mathop{\lim }}\,\dfrac{1}{n}\sum\limits_{r=1}^{2n}{\ln \left( 1+\dfrac{r}{n} \right)}\]
Lastly, we will use the ILATE formula of the integral to integrate where we use \[\int{\left( u.v \right)dx=}u.\int{vdx}-\int{\left[ \dfrac{du}{dx}\left( \int{vdx} \right) \right]dx}\] to get our required answer.

Complete step by step answer:

We have to find the limit of \[{{\left( \dfrac{\left( n+1 \right)\left( n+2 \right).....\left( 3n \right)}{{{n}^{2n}}} \right)}^{\dfrac{1}{n}}}\] as \[n\to \infty .\] Let us assume the limit as y. So, we get,
\[y=\underset{n\to \infty }{\mathop{\lim }}\,{{\left( \dfrac{\left( n+1 \right)\left( n+2 \right).....\left( 3n \right)}{{{n}^{2n}}} \right)}^{\dfrac{1}{n}}}\]
Now we take log on both the sides, we will get,
\[\ln y=\underset{n\to \infty }{\mathop{\lim }}\,\ln {{\left( \dfrac{\left( n+1 \right)}{n}\dfrac{\left( n+2 \right)}{n}......\dfrac{\left( n+2n \right)}{n} \right)}^{\dfrac{1}{n}}}\]
As, \[\log {{a}^{b}}=b\log a,\] so we get,
\[\ln y=\underset{n\to \infty }{\mathop{\lim }}\,\dfrac{1}{n}\ln \left( \dfrac{\left( n+1 \right)}{n}\dfrac{\left( n+2 \right)}{n}......\dfrac{\left( n+2n \right)}{n} \right)\]
We know that,
\[\log \left( abc \right)=\log a+\log b+\log c\]
So, we get,
\[\ln y=\underset{n\to \infty }{\mathop{\lim }}\,\dfrac{1}{n}\left[ \ln \left( 1+\dfrac{1}{n} \right)+\ln \left( 1+\dfrac{2}{n} \right).....+\ln \left( 1+\dfrac{2n}{n} \right) \right]\]
We can clearly see that we are getting a summation of \[\log \left( 1+\dfrac{r}{n} \right)\] here. So, we get,
\[\ln y=\underset{n\to \infty }{\mathop{\lim }}\,\dfrac{1}{n}\sum\limits_{r=1}^{2n}{\ln \left( 1+\dfrac{r}{n} \right)}\]
Let, \[\dfrac{r}{n}=x,\] so, \[\dfrac{dr}{n}=dx.\]
And, when r = 1, so, \[x=\dfrac{1}{n}.\]
As, \[n\to \infty ,\] so, x = 0.
Therefore, r = 1 means, x = 0.
And when r = 2n, so, \[x=\dfrac{2n}{n}=2.\]
So,
\[\ln y=\underset{n\to \infty }{\mathop{\lim }}\,\dfrac{1}{n}\sum\limits_{r=1}^{2n}{\ln \left( 1+\dfrac{r}{n} \right)}\]
Becomes
\[\ln y=\int\limits_{0}^{2}{\ln \left( 1+n \right)dx}\]
Now we use ILATE, so, our first function is u = ln (1 + n) and second function is v = 1.
We know that,
\[\int\limits_{a}^{b}{\left( u.v \right)dx}=u.\int\limits_{a}^{b}{vdx}-\int\limits_{a}^{b}{\left[ \dfrac{du}{dx}\int{vdx} \right]dx}\]
So applying this on \[\int\limits_{0}^{2}{\ln \left( 1+n \right)dn}\] with u = ln (1 + n) and v = 1. So, we get,
\[\int\limits_{0}^{2}{\ln \left( 1+n \right)dn=\ln \left( 1+n \right)\int\limits_{0}^{2}{dn}}-\int\limits_{0}^{2}{\left[ \dfrac{d\left( \ln \left( 1+n \right) \right)}{dn}\int{dn} \right]}dn\]
\[\int\limits_{0}^{2}{\ln \left( 1+n \right)dn=\left[ n.\ln \left( 1+n \right) \right]_{0}^{2}}-\int\limits_{0}^{2}{\dfrac{n}{1+n}dn}......\left( i \right)\]
Let, \[{{I}_{1}}=\int\limits_{0}^{2}{\dfrac{n}{1+n}dn}\]
\[\Rightarrow {{I}_{1}}=\int\limits_{0}^{2}{\dfrac{n+1-1}{1+n}dn}\]
\[\Rightarrow {{I}_{1}}=\int\limits_{0}^{2}{1-\dfrac{1}{1+n}dn}\]
\[\Rightarrow {{I}_{1}}=\left( n \right)_{0}^{2}-\left[ \ln \left( 1+n \right) \right]_{0}^{2}......\left( ii \right)\]
Using (i) and (ii), we get,
\[\int\limits_{0}^{2}{\ln \left( 1+n \right)dn}=\left[ n\ln \left( 1+n \right) \right]_{0}^{2}-\left[ \left( n \right)_{0}^{2}-\left( \ln \left( 1+n \right) \right)_{0}^{2} \right]\]
\[\Rightarrow \int\limits_{0}^{2}{\ln \left( 1+n \right)dn}=\left( 2\ln 3-0 \right)-\left[ \left( 2-0 \right)-\left( \ln 3-0 \right) \right]\]
Simplifying, we get,
\[\Rightarrow \int\limits_{0}^{2}{\ln \left( 1+n \right)dn}=2\ln 3+\ln 3-2\]
\[\Rightarrow \int\limits_{0}^{2}{\ln \left( 1+n \right)dn}=3\ln 3-2\]
We know that, ln e = 1. So,
\[\ln y=3\ln 3-2\ln e\]
We know that, \[\log {{a}^{b}}=b\log a.\]
So,
\[\ln y=\ln {{3}^{3}}-\ln {{e}^{2}}\]
As, \[\ln \left( \dfrac{a}{b} \right)=\ln a-\ln b,\] we get,
\[\Rightarrow \ln y=\ln \left( \dfrac{{{3}^{3}}}{{{e}^{2}}} \right)\]
Comparing both the sides, we get,
\[y=\dfrac{{{3}^{3}}}{{{e}^{2}}}=\dfrac{27}{{{e}^{2}}}\]
So, our limit is \[\dfrac{27}{{{e}^{2}}}.\]
Hence, the right option is (b).

Note:
 Remember that \[\int{\dfrac{1}{1+n}dn}=\log \left( 1+n \right)\] because we took 1 + n as t and then differentiate both the sides, we get, dx = dt. So, \[\int{\dfrac{1}{1+n}dn}=\int{\dfrac{1}{t}dt}\] and \[\int{\dfrac{1}{t}dt}=\log t.\] So, we get, \[\int{\dfrac{1}{1+n}dn}=\log t.\] Putting t back as 1 + n, we get, \[\int{\dfrac{1}{x}dx}=\log \left( 1+x \right)\] but \[\int{\dfrac{1}{1-x}dx}\ne \log \left( 1-x \right)\] because if we take 1 – x as t, we get, dx = – dt when we give
\[\int{\dfrac{1}{1-x}dx}=\int{\dfrac{1}{t}\left( -dt \right)}\]
\[\Rightarrow -\log t=-\log \left( 1-x \right)\]