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

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.

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

• Quite an elegant solution. Just out of curiosity, was I supposed to know the last identity from undegrad Real Analysis (only for single variable functions)? – Feynman_00 Oct 24 at 7:48
• @Feynman_00 We can prove it by the definition of the Riemann's integral. If you want I am ready to show. – Michael Rozenberg Oct 24 at 7:56
• I'd be very grateful if you did (if the rules allow it). – Feynman_00 Oct 24 at 7:59
• @Feynman_00 I added something. See now. – Michael Rozenberg Oct 24 at 8:06
• I understand. The only integral definition I was given was Darboux integral that only involves supremum of inferior sum and infimum of superior sum. Riemann's limit definition is quite useful, I'll study something about it. – Feynman_00 Oct 24 at 9:36

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.

• Each terms tends to $1$ therefore it is not so clear how to determine the limit. Maybe you could add something more on that. – user Oct 24 at 8:08
• @user All factors, except for the first, get arbitrarily close to $1$ at $n=\infty.$ Obviously, the first becomes unbounded, since the exponent is harmonic. – Allawonder Oct 24 at 8:14
• Ah ok, I didn't read properly your first step. Now it is clear to me. Thanks – user Oct 24 at 8:19

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

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

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

• I like this solution as well. Once again, as I said above those sums and big $\mathcal{O}$ are quite new to me, so (just out of curiosity) was I supposed to know that from an undegrad real analysis course? – Feynman_00 Oct 24 at 7:57
• @Feynman_00 Yes in same cases it is a very useful notation to simplify the presentation and I suppose it should be covered in any real analysis course. – user Oct 24 at 8:01
• @Feynman_00 In this case, following the idea from other answers, we can obtain the result by simpler considerations. I've added a solution using that way. – user Oct 24 at 8:06