How do you evaluate the limit of this sequence involving n-th roots?

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 1

$$\prod_{k=n}^{2n}\sqrt[k]k=e^{\sum\limits_{k=n}^{2n}\frac{\ln{k}}{k}}>e^{\ln{n}\sum\limits_{k=n}^{2n}\frac{1}{k}}=n\cdot e^{\sum\limits_{k=n}^{2n}\frac{1}{k}}\rightarrow+\infty$$ because $$\lim_{n\rightarrow+\infty}\sum\limits_{k=n}^{2n}\frac{1}{k}=\lim_{n\rightarrow+\infty}\left(\frac{1}{1+\frac{0}{n}}+\frac{1}{1+\frac{1}{n}}+...+\frac{1}{1+\frac{n}{n}}\right)\frac{1}{n}=$$ $$=\int\limits_0^1\frac{1}{1+x}dx=\ln(1+x)|_0^1=\ln2.$$

Answer 2

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 $\sum \frac{\log n}{n}$

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

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

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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

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

Answer 3

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.