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 $$\lim\_{n\rightarrow+\infty}\sum\limits\_{k=n}^{2n}\frac{1}{k}=\lim\_{n\rightarrow+\infty}\left(\frac{1}{1+\frac{0}{n}}+\frac{1}{1+\frac{1}{n}}+...+\frac{1}{1+\frac{n}{n}}\right)\frac{1}{n}=$$
$$=\int\limits\_0^1\frac{1}{1+x}dx=\ln(1+x)|\_0^1=\ln2.$$
We have that

$$\sqrt[n]{n}\cdot\sqrt[n+1]{n+1}...\sqrt[2n]{2n}=e^{\sum\_{k=n}^{2n}\frac{\log k}{k}}$$

and by this result

* [Big $\mathcal{O}$ Notation question while estimating $\sum \frac{\log n}{n}$](https://math.stackexchange.com/q/13300/505767)

we obtain

$$\sum\_{k=n}^{2n}\frac{\log k}{k}=\sum\_{k=1}^{2n}\frac{\log k}{k}-\sum\_{k=1}^{n-1}\frac{\log k}{k} =\frac{(\log 2n)^2}{2}-\frac{(\log (n-1))^2}{2} + O\left(\frac{\log(n)}{n}\right)$$

with

$$(\log 2n)^2-(\log (n-1))^2=\left(\log \left(\frac{2n}{n-1}\right)\right)(\log (2n(n-1)) \to \infty$$

---

More simply we have

$$\sum\_{k=n}^{2n}\frac{\log k}{k} \ge n \cdot \frac{\log (2n)}{2n}=\frac{\log (2n)}{2} $$

and therefore

$$e^{\sum\_{k=n}^{2n}\frac{\log k}{k}} \ge e^{\frac{\log (2n)}{2}}\to \infty$$