<s>[INST] Evaluating $\lim_{n\to\infty} \left(\prod_{r=0}^n \binom{n}{r}\right)^\frac{1}{n(n+1)}$ 
Problem statement:
If
$$ A = \lim_{n\to\infty} \left(\prod_{r=0}^n \binom{n}{r}\right)^\frac{1}{n(n+1)}, $$
Find $A^2$.
Solution:1)
\begin{align*}
\prod_{k=0}^{n} \binom{n}{k}
&= \prod_{k=1}^{n} \frac{n!}{k!(n-k)!}
= \frac{(n!)^{n+1}}{(1! \cdot 2!\cdots n!)^2} \\
&= \prod_{k=1}^{n} (n+1-k)^{n+1-2k} \\
&= \prod_{k=1}^{n} \left(\frac{n+1-k}{n+1}\right)^{n+1-2k},
\end{align*}
since $\sum_{k=1}^{n} (n+1-2k) = 0$. Taking log and limit, we get
\begin{align*}
& \frac{1}{n}\sum_{k=1}^{n}\left( 1 - \frac{2k}{n+1} \right) \ln\left( 1 - \frac{k}{n+1} \right) \\
&\to \int_{0}^{1} (1 - 2x)\log(1-x) \, \mathrm{d}x
= \frac{1}{2}.
\end{align*}
Source: FITJEE AITS 2020 FT-8 Paper 1 of JEE Advanced question 53

Could anybody explain the solution?
I tried taking log both sides to bring the power down but got stuck on evaluating $$\sum_{r=0}^n \ln\binom{n}{r}$$ further I think the answer should be $e$ instead of $0.5$ which is given.
 [/INST] Let $A_n = \left(\prod_{r=0}^n \binom{n}{r}\right)^\frac{1}{n(n+1)}$ denote the expression before the limit is taken. Then
\begin{align*}
\log A_n
&= \frac{1}{n(n+1)} \log \left[\prod_{r=0}^n \binom{n}{r}\right] \tag{1} \\
&= \frac{1}{n(n+1)} \sum_{r=0}^{n} \log \binom{n}{r} \tag{2} \\
&= \frac{1}{n(n+1)} \sum_{r=0}^{n} \left[ \log n! - \log r! - \log(n-r)! \right] \tag{3} \\
&= \frac{1}{n(n+1)} \left[ (n+1) \log n! - 2 \sum_{r=1}^{n} \log r! \right] \tag{4} \\
&= \frac{1}{n(n+1)} \left[ (n+1) \left( \sum_{k=1}^{n} \log k \right) - 2 \sum_{r=1}^{n} \sum_{k=1}^{r} \log k \right] \tag{5} \\
&= \frac{1}{n(n+1)} \left[ (n+1) \left( \sum_{k=1}^{n} \log k \right) - 2 \sum_{k=1}^{n} \sum_{r=k}^{n} \log k \right] \tag{6}  \\
&= \frac{1}{n(n+1)} \left[ (n+1) \left( \sum_{k=1}^{n} \log k \right) - \sum_{k=1}^{n} 2(n+1-k) \log k \right] \tag{7} \\
&= \frac{1}{n(n+1)} \sum_{k=1}^{n} (2k-n-1) \log k \\
&= \frac{1}{n} \sum_{k=1}^{n} \left(\frac{2k}{n+1}-1\right) \log k \tag{8}
\end{align*}
Note:

*

*$\text{(1)}$ : $\log(a^b) = b \log a$ for $a, b > 0$.

*$\text{(2)}$ : $\log(ab) = \log a + \log b$.

*$\text{(3)}$ : $\binom{n}{r} = \frac{n!}{r!(n-r)!}$

*$\text{(4)}$ : $\sum_{r=0}^{n} \log (n-r)! = \log n! + \log(n-1)! + \cdots + \log 1! + \log 0! = \sum_{r=0}^{n} \log r!$. (Or, substitute $r' = n - r$.)

*$\text{(5)}$ : $k! = 1 \cdot 2 \cdots (n-1) \cdot n $ and then use the property of $\log$.

*$\text{(6)}$ : Swapping the order of summation; the double sum runs over $(r, k)$ where $1 \leq k \leq r \leq n$. This is the same as $1 \leq k \leq n$ and $ k \leq r \leq n$.

*$\text{(7)}$ : The inner sum is $\sum_{r=k}^{n} \log k = (\log k) \sum_{r=k}^{n} 1 = (\log k)(n+1-r) $.

Now, $\text{(8)}$ almost looks like a Riemann sum, and it will be indeed a Riemann sum if $\log k$ were $\log \frac{k}{n+1}$ instead. So, it is natural to replace $\log k$ by $\log\frac{k}{n+1}$ and then examine how the difference behaves. Proceeding,
\begin{align*}
\log A_n
&= \frac{1}{n} \sum_{k=1}^{n} \left(\frac{2k}{n+1}-1\right) \left[ \log\left(\frac{k}{n+1}\right) + \log (n+1) \right] \\
&= \frac{1}{n} \sum_{k=1}^{n} \left(\frac{2k}{n+1}-1\right) \log\left(\frac{k}{n+1}\right) + \frac{\log (n+1)}{n(n+1)} \sum_{k=1}^{n} (2k - n - 1). \tag{9}
\end{align*}
However, we have
$$ \sum_{k=1}^{n} (2k - n - 1)
= n(n+1) - n(n+1)
= 0. $$
So the second sum in $\text{(9)}$ is zero and
\begin{align*}
\log A_n
&= \frac{1}{n} \sum_{k=1}^{n} \left(\frac{2k}{n+1}-1\right) \log\left(\frac{k}{n+1}\right) \\
&\to \int _{0}^{1} (2x - 1)\log x \, \mathrm{d}x \\
&= \left[ x - \frac{x^2}{2} + (x^2 - x) \log x \right]_{0}^{1} \\
&= \frac{1}{2}.
\end{align*}
Therefore $ \log A = \frac{1}{2} $ and hence $A^2 = e^{2\log A} = e$.
</s>