Let ${{P}_{n}}=\sqrt[n]{\dfrac{\left( 3n \right)!}{\left( 2n \right)!}}\left( n=1,2,3...... \right)$ then find $\underset{n\to \infty }{\mathop{\lim }}\,\dfrac{{{P}_{n}}}{n}$

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Hint: Simplify the given limit then apply limit as a sum concept of integral chapter.

Hence, we have equation;
${{P}_{n}}={{\left( \dfrac{3n!}{2n!} \right)}^{\dfrac{1}{n}}}\left( n=1,2,3...... \right)$
And we have to find $\underset{n\to \infty }{\mathop{\lim }}\,\dfrac{{{P}_{n}}}{n}$i.e.
\[\underset{n\to \infty }{\mathop{\lim }}\,\dfrac{1}{n}{{\left( \dfrac{\left( 3n \right)!}{\left( 2n \right)!} \right)}^{\dfrac{1}{n}}}\]
Let us suppose the above relation as
$y=\underset{n\to \infty }{\mathop{\lim }}\,\dfrac{1}{n}{{\left( \dfrac{3n!}{2n!} \right)}^{\dfrac{1}{n}}}.........\left( 1 \right)$
We cannot directly put a limit to the given function as $n$ is not defined.
We need to first simplify the given relation.
$y=\underset{n\to \infty }{\mathop{\lim }}\,\dfrac{1}{n}{{\left( \dfrac{\left( 3n \right)\left( 3n-1 \right)\left( 3n-2 \right)........1}{\left( 2n \right)\left( 2n-1 \right)\left( 2n-2 \right)........1} \right)}^{\dfrac{1}{n}}}$
Let us write $\left( 3n \right)!$ in reverse order
$y=\underset{n\to \infty }{\mathop{\lim }}\,\dfrac{1}{n}{{\left( \dfrac{1.2.3................3n}{\left( 2n \right)\left( 2n-1 \right)\left( 2n-2 \right)..........2.1} \right)}^{\dfrac{1}{n}}}$
As we know that $\left( 2n<3n \right)$ ; hence $2n$ will come in between $1\And 3n$ . Hence, we can rewrite the above relation as;
\[y=\underset{n\to \infty }{\mathop{\lim }}\,\dfrac{1}{n}{{\left( \dfrac{\left( 2n \right)\left( 2n+1 \right)\left( 2n+2 \right)......3n}{\left( 2n \right)\left( 2n-1 \right)\left( 2n-2 \right).........2.1} \right)}^{\dfrac{1}{n}}}\]
Now, by simplifying the above equation, we can observe that $\left( 2n \right)!$ is common in numerator and denominator. Hence, we can cancel out them and we get the above equation as;
$y=\underset{n\to \infty }{\mathop{\lim }}\,{{\left( \dfrac{\left( 2n+1 \right)\left( 2n+2 \right)\left( 2n+3 \right)........3n}{{n}^{n}} \right)}^{\dfrac{1}{n}}}$
Now, let us take the denominator $'n'$ to the bracket of numerators. $'n'$ will be converted to ${{n}^{n}}$ as power of numerator is $\dfrac{1}{n}$.
Therefore, $y$can be written as;
  & y=\underset{n\to \infty }{\mathop{\lim }}\,{{\left( \dfrac{\left( 2n+1 \right)\left( 2n+2 \right).....3n}{{{n}^{n}}} \right)}^{\dfrac{1}{n}}} \\
 & \because {{n}^{n}}=n.n.n.n....\left( n\text{ number of times} \right) \\
 & y=\underset{n\to \infty }{\mathop{\lim }}\,{{\left( \left( \dfrac{2n+1}{n} \right)\left( \dfrac{2n+2}{n} \right)......\left( \dfrac{3n}{n} \right) \right)}^{\dfrac{1}{n}}} \\
 & y=\underset{n\to \infty }{\mathop{\lim }}\,{{\left( \left( 2+\dfrac{1}{n} \right)\left( 2+\dfrac{2}{n} \right)\left( 2+\dfrac{3}{n} \right).........\left( 2+\dfrac{n}{n} \right) \right)}^{\dfrac{1}{n}}}........\left( 2 \right) \\
We cannot put a limit to the equation $\left( 2 \right)$ as well, because the limit is not defined till now.
So, taking log to both sides of equation $\left( 2 \right)$
$\log y=\underset{x\to \infty }{\mathop{\lim }}\,\log {{\left( \left( 2+\dfrac{1}{n} \right)\left( 2+\dfrac{2}{n} \right)\left( 2+\dfrac{3}{n} \right)......\left( 2+\dfrac{n}{n} \right) \right)}^{\dfrac{1}{n}}}$
As we know property of logarithm as
  & \log {{a}^{m}}=m\log a \\
 & \log ab=\log a+\log b \\
From the above relation (later one), we can write this equation for $n$ values as;
$\log \left( {{x}_{1}}{{x}_{2}}......{{x}_{n}} \right)=\log {{x}_{1}}+\log {{x}_{2}}+\log {{x}_{3}}+.......\log {{x}_{n}}$
Hence, using the above properties of logarithm with the equation $'\log y'$ we get
  & \log =\underset{n\to \infty }{\mathop{\lim }}\,\dfrac{1}{n}\log \left( 2+\dfrac{1}{n} \right)\left( 2+\dfrac{2}{n} \right)\left( 2+\dfrac{3}{n} \right).....\left( 2+\dfrac{n}{n} \right) \\
 & \log =\underset{n\to \infty }{\mathop{\lim }}\,\dfrac{1}{n}\left( \log \left( 2+\dfrac{1}{n} \right)+\log \left( 2+\dfrac{2}{n} \right)+\log \left( 2+\dfrac{3}{n} \right)+............\log \left( 2+\dfrac{n}{n} \right) \right)........\left( 3 \right) \\
Now, as we have learnt in integration that we can convert any infinite series with summation to integral by using limit as a sum concept.
So, let us first write equation $\left( 3 \right)$ in form of limit as sum:
$\log y=\underset{n\to \infty }{\mathop{\lim }}\,\dfrac{1}{n}\sum\limits_{r=1}^{n}{\log \left( 2+\dfrac{r}{n} \right)............\left( 4 \right)}$
We have learnt that if any summation given as;
$\underset{n\to \infty }{\mathop{\lim }}\,\dfrac{1}{n}\sum\limits_{r=1}^{n}{f\left( a+\dfrac{r}{n} \right)}$
Then we can convert this series into integral by replacing
  & \dfrac{1}{n}\text{ to }dx \\
 & \dfrac{r}{n}\text{ to }x \\
Lower limit = put $n\to \infty \text{ with }\left( \dfrac{r}{n} \right)$
Upper limit = put maximum value of$\text{ }r\text{ in }\left( \dfrac{r}{n} \right)$
Hence, we can convert equation $\left( 4 \right)$ as
  & \log y=\underset{n\to \infty }{\mathop{\lim }}\,\dfrac{1}{n}\sum\limits_{r=1}^{n}{\log \left( 2+\dfrac{r}{n} \right)} \\
 & \text{upper limit=}1\left( \dfrac{r}{n}=\dfrac{n}{n}=1 \right) \\
 & \text{lower limit=0}\left( {}_{n\to \infty }\dfrac{r}{n}=0 \right) \\
$\log y=\int_{0}^{1}{\log \left( 2+x \right)dx.............\left( 5 \right)}$
Above integration can be solved by integration by parts which can be expressed as, if we have two functions in multiplication as $\int{f\left( x \right)g\left( x \right)dx}$may be different to each other (Example: one trigonometric and another algebraic) then we can calculate the integration by using relation$\int{f\left( x \right)g\left( x \right)dx}=f\left( x \right)\int{g\left( x \right)dx-\int{f'\left( x \right)\int{g\left( x \right)dx.dx.........\left( 6 \right)}}}$
Applying integration by parts as expressed in equation $\left( 6 \right)$ with equation $\left( 5 \right)$, as
$f\left( x \right)=1\And g\left( x \right)=\log \left( 2+x \right)$.
  & \log y=\int_{0}^{1}{1.\log \left( 2+x \right)dx} \\
 & \log y=\log \left( 2+x \right)\int_{0}^{1}{1dx}-\int_{0}^{1}{\dfrac{d}{dx}\left( \log \left( 2+x \right)\int{1dx} \right)dx} \\
We know
  & \int{{{x}^{n}}=\dfrac{{{x}^{n+1}}}{n+1}} \\
 & \dfrac{d}{dx}\log x=\dfrac{1}{x} \\
  & \log y=x\log \left( 2+x \right)\left| \begin{matrix}
   1 \\
   0 \\
\end{matrix} \right.-\int_{0}^{1}{\dfrac{1}{2+x}\left( x \right)dx} \\
 & \log y=\left( 1\log 3-0 \right)-\int_{0}^{1}{\dfrac{x}{2+x}dx} \\
 & \log y=\log 3-\int_{0}^{1}{\left( \dfrac{\left( 2+x \right)-2}{2+x} \right)}dx \\
 & \log y=\log 3-\int_{0}^{1}{\left( 1-\dfrac{2}{2+x} \right)}dx \\
 & \log y=\log 3=\int_{0}^{1}{1dx+2}\int_{0}^{1}{\dfrac{1}{2+x}dx} \\
 & \int{\dfrac{1}{x}dx}=\log x={{\log }_{e}}x \\
 & \log y=\log 3-x\left| \begin{matrix}
   1 \\
   0 \\
\end{matrix} \right.+2\log \left( 2+x \right)\left| \begin{matrix}
   1 \\
   0 \\
\end{matrix} \right. \\
 & \log y=\log 3-1+2\log 3-2\log 2 \\
 & \log y=3\log 3-2\log 2-1..............\left( 7 \right) \\
 We know that,
$\log {{a}^{m}}=m\log a$ or vice versa
  & \log y=\log {{3}^{3}}-\log {{2}^{2}}-1 \\
 & \log y=\log 27-\log 4-1.............\left( 8 \right) \\
We also have property of logarithm as
  & \log a-\log b=\log \left( \dfrac{a}{b} \right) \\
 & \log {{a}^{a}}=1 \\
As, equation $\left( 8 \right)$ have log with (default) base $e$ , so we can replace $1\text{ to }\log {{e}^{e}}$
  & \log {{e}^{y}}=\log e\dfrac{27}{4}-\log {{e}^{e}} \\
 & \log y=\log \dfrac{27}{4e} \\
$y=\dfrac{27}{4e}$ by comparison.

Note: This question does not belong to limit and differentiability chapter but from limit as a sum concept of integrals. It is the key point of this question to get a solution.
One can go wrong while calculating lower limit and upper limit which is defined as
Lower limit$=$ put $n\to \infty \text{ with }\left( \dfrac{r}{n} \right)$
Upper limit$=$ put maximum value of$\text{ }r\text{ in }\left( \dfrac{r}{n} \right)$
Integration of $\log \left( 2+x \right)$ is also an important step. As we need to multiply by $1$ for applying integration by parts.