I need to find where this sequence tends to: $$\frac{1}{n} \sqrt[n]{n(n+1)(n+2)...(2n)}$$
My answer is $2$, by using a trick for the $n^{th}$ root: I take the inside and find where $\dfrac{A_{n+1}}{A_n}$ goes, and this is where the original sequence goes.
$$ a_n = \frac{1}{n} \sqrt[n]{n(n+1)(n+2)\cdots (2n)} = \sqrt[n]{1(1+\frac{1}{n})(1+\frac{2}{n})\cdots (1+\frac{n}{n})}. $$ $$ \log a_n = \frac{1}{n} \{ \log 1 + \log (1+\frac{1}{n}) + \log (1+\frac{2}{n}) + \cdots + \log (1+\frac{n}{n}) \}. $$
We know the above equations. Hence,
$$ \lim\limits_{n \rightarrow \infty} \log a_n = \int_1^2 \log x dx = [x\log x -x ]_1^2 = 2\log 2 - 1 = \log \frac{4}{e}. $$
Thus, the limit of $a_n$ is $\frac{4}{e}$.
By ratio root criterion we have
$$a_n=\frac{n(n+1)(n+2)...(2n)}{n^n} \quad b_n=\frac{1}{n} \sqrt[n]{n(n+1)(n+2)...(2n)}=\sqrt[n] a_n$$
then
$$\frac{a_{n+1}}{a_n}=\frac{(n+1)(n+2)...(2n+2)}{n(n+1)(n+2)...(2n)}\frac{n^n}{(n+1)^{n+1}}=\frac{(2n+2)(2n+1)}{n(n+1)}\frac{1}{\left(1+\frac1n\right)^n}\to \frac 4 e$$
which implies that $b_n \to \frac 4 e$.
By Stirling's formula, \begin{align} \ln\left[(n+1)(n+2)\cdots(2n)\right] &=\ln (2n)!-\ln n!=(2n+1/2)\ln(2n)-(n+1/2)\ln n-n+o(1)\\ &=(2n+1/2)\ln 2+n\ln n-n+o(1) \end{align} and so $$\ln\left[n(n+1)(n+2)\cdots(2n)\right] =(2n+1/2)\ln 2+(n+1)\ln n-n+o(1).$$ Then $$\ln\sqrt[n]{n(n+1)(n+2)\cdots(2n)}=2\ln 2+\ln n-1+o(1)$$ and $$\ln\frac{\sqrt[n]{n(n+1)(n+2)\cdots(2n)}}n=2\ln 2-1+o(1).$$ We conclude that your sequence converges to $4/e$.
\frac{numerator}{denominator}and\sqrt[n]{expression}– Yuriy S Nov 10 '18 at 15:01