Evaluate $\lim_{n \rightarrow \infty } \frac {[(n+1)(n+2)\cdots(n+n)]^{1/n}}{n}$ Evaluate $$\lim_{n \rightarrow \infty~} \dfrac {[(n+1)(n+2)\cdots(n+n)]^{\dfrac {1}{n}}}{n}$$ using Cesáro-Stolz theorem.
I know there are many question like this, but i want to solve it using Cesáro-Stolz method and no others.
I took log and applied Cesáro-Stolz, I get $$\log{2}+n\log\cfrac{n}{n+1}$$
Which gives me answer as $\frac{2}{e}$ .
But answer is $\frac{4}{e}$. Could someone help?.
Edit: 
On taking log, 
$$\lim_{n \to \infty} \frac{-n\log n + \sum\limits_{k=1}^{n} \log \left(k+n\right)}{n} \\= \lim_{n \to \infty} \left(-(n+1)\log (n+1) + \sum\limits_{k=1}^{n+1} \log \left(k+n\right)\right) - \left(-n\log n + \sum\limits_{k=1}^{n} \log \left(k+n\right)\right) \\ = \lim_{n \to \infty} \log \frac{2n+1}{n+1} - n\log \left(1+\frac{1}{n}\right) = \log 2 - 1$$
Which gives $2/e$
 A: After took log, we have
$$\frac{\sum _{i=1}^n \log (i+n)-n \log (n)}{n}$$
Applied cesaro stolez, we have
\begin{align}
&
\left(\sum _{i=1}^{n+1} \log (i+n+1)-(n+1) \log (n+1)\right)-\left(\sum _{i=1}^n \log (i+n)-n \log (n)\right) \\&= 
\left(\sum _{i=2}^{n+2} \log (i+n)-(n+1) \log (n+1)\right)-\left(\sum _{i=1}^n \log (i+n)-n \log (n)\right)\\&=
(\log (2 n+1)+\log (2 n+2)-(n+1) \log (n+1))-(\log (n+1)-n \log (n)) \\ &=
n \log \left(\frac{n}{n+1}\right)+\log \left(\frac{(2 n+1) (2 n+2)}{(n+1)^2}\right) \\ &\rightarrow -1+\log (4)
\end{align}
A: Let $a_n=\sum_{i=1}^{n} \log\left(1+\frac{i}n\right)$ be the numerator of the logarithm, with denominator $b_n=n.$
The key is that the first term $\log(1+1/n)$ of $a_n$ doesn't cancel with any of the terms $\log(1+i/(n+1)).$ It alone is subtracted, so, while there are $n$ occurrences of $-\log(1+1/n)$, there are still two more terms for $a_{n+1}.$ So you get:
$$a_{n+1} - a_n = \left(\frac{2n+1}{n+1}\right)+\log\left(\frac{2n+2}{n+1}\right)-n\log(1+1/n)$$

Basically, the "cancellation" happens when we have $$\log\left(1+\frac{i}{n+1}\right)-\log\left(1+\frac{i+1}{n}\right)=
\log\left(\frac{n}{n+1}\right)$$
While we can assume there is a $i=0$ in $a_{n+1},$ since it adds $0=\log 1$ to $a_{n+1},$ that means there are $n+2$ values of $i$ in $\sum_{i=0}^{n+1},$ and hence there is not cancellation of the $i=n$ and $i=n+1$ terms from $a_{n+1}.$
