Question: This is the limit:
\begin{equation}
\lim\_{n\to+\infty}\sqrt[n]{n}\cdot\sqrt[n+1]{n+1}...\sqrt[2n]{2n}
\end{equation}
I tried to rearrange the terms to apply the geometric mean theorem but my attempt was not successful. Any way to solve this will be fine.

Answer: 
$$\prod\_{k=n}^{2n}\sqrt[k]k=e^{\sum\limits\_{k=n}^{2n}\frac{\ln{k}}{k}}>e^{\ln{n}\sum\limits\_{k=n}^{2n}\frac{1}{k}}=n\cdot e^{\sum\limits\_{k=n}^{2n}\frac{1}{k}}\rightarrow+\infty$$ because $$\lim\_{n\rightarrow+\infty}\sum\limits\_{k=n}^{2n}\frac{1}{k}=\lim\_{n\rightarrow+\infty}\left(\frac{1}{1+\frac{0}{n}}+\frac{1}{1+\frac{1}{n}}+...+\frac{1}{1+\frac{n}{n}}\right)\frac{1}{n}=$$
$$=\int\limits\_0^1\frac{1}{1+x}dx=\ln(1+x)|\_0^1=\ln2.$$
Factor out $n$ from each of the $n$ factors on the right, to obtain $$n^{1/n+1/(n+1)+1/(n+2)+\cdots+1/2n}\left(1+\frac1 n\right)^{1/(n+1)}\left(1+\frac1 n\right)^{1/(n+2)}\cdots\left(1+\frac1 n\right)^{1/2n},$$ which makes the limit at $n=\infty$ clear.