<s>[INST] What value does $\frac{1}{n} \sqrt[n]{n(n+1)(n+2)\cdots(2n)}$ tend to? 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.
 [/INST] 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 $b_n \to \frac 4 e$.
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