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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.

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

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  • $\begingroup$ 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)? $\endgroup$ – Feynman_00 Oct 24 at 7:48
  • $\begingroup$ @Feynman_00 We can prove it by the definition of the Riemann's integral. If you want I am ready to show. $\endgroup$ – Michael Rozenberg Oct 24 at 7:56
  • $\begingroup$ I'd be very grateful if you did (if the rules allow it). $\endgroup$ – Feynman_00 Oct 24 at 7:59
  • $\begingroup$ @Feynman_00 I added something. See now. $\endgroup$ – Michael Rozenberg Oct 24 at 8:06
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    $\begingroup$ 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. $\endgroup$ – Feynman_00 Oct 24 at 9:36
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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.

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  • $\begingroup$ Each terms tends to $1$ therefore it is not so clear how to determine the limit. Maybe you could add something more on that. $\endgroup$ – user Oct 24 at 8:08
  • $\begingroup$ @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. $\endgroup$ – Allawonder Oct 24 at 8:14
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    $\begingroup$ Ah ok, I didn't read properly your first step. Now it is clear to me. Thanks $\endgroup$ – user Oct 24 at 8:19
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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

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

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  • $\begingroup$ 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? $\endgroup$ – Feynman_00 Oct 24 at 7:57
  • $\begingroup$ @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. $\endgroup$ – user Oct 24 at 8:01
  • $\begingroup$ @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. $\endgroup$ – user Oct 24 at 8:06

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