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Let $(a_{n})\subset\mathbb{R}$ be such that $0\le a_{n+m}\le a_{n}a_{m}$ for all $n,m\in\mathbb{N}$. I want to show that $(\sqrt[n]{a_{n}})_{n}$ converges.

Certainly, I thought of trying to ratio test, but this only yielded:

$$\left|\frac{\sqrt[n+1]{a_{n+1}}}{\sqrt[n]{a_{n}}}\right|=\left|\frac{a_{n+1}}{a_{n}}\right|\le\left|\frac{a_{n}a_{1}}{a_{n}}\right|=|a_{1}|\xrightarrow{n\to\infty}|a_{1}|$$

which is inconclusive. Perhaps someone could offer a better suggestion?

Edit: See the comments. I did some sloppy (i.e. wrong) calculations.

Jason Born
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  • That's probably just me being tired, but how do you get the first equality? – Clement C. Oct 22 '16 at 18:38
  • @ClementC. No, it seems that I was in fact tired since we have in fact that $$\begin{aligned}\left|\frac{(a_{n+1})^{\frac{1}{n+1}}}{(a_{n})^{\frac{1}{n}}}\right|&=|(a_{n+1})^{\frac{1}{n+1}}(a_{n})^{-\frac{1}{n}}| \ &\le|(a_{1})^{\frac{1}{n+1}}(a_{n})^{\frac{1}{n+1}}(a_{n})^{-\frac{1}{n}}| \ &=|(a_{1})^{\frac{1}{n+1}}(a_{n})^{-\frac{1}{n(n+1)}}|\end{aligned}$$ and if we take $n\to\infty$ then we just end up with 1, which is no good (i.e. the test was inconclusive). – Jason Born Oct 22 '16 at 18:52
  • I'm going to nitpick on a tangent. What is the purpose of the $n$ index in $(\sqrt[n]{a_n})_{\rightarrow n \leftarrow}$? I don't see that it is nescessary or in this case that it actually means anything or even makes sense. Okay, that was a diversion. It doesn't actually affect the problem but ... – fleablood Oct 22 '16 at 19:41

1 Answers1

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Partial answer. From $0\le a_{n+m}\le a_{n}a_{m}$ we get $a_n \le (a_1)^n$ therefore $\sqrt[n]{a_{n}} \le a_1$ so the sequence $(\sqrt[n]{a_{n}})_{n}$ is bounded.

My answer only proves the sequence is bounded, not convergent. Therefore I don't think accepting this answer is appropriate.

  • I don't see how $a_{n}\le (a_{1})^{n}$ follows. In order to use that inequality and have both $a_{n}$ and $a_{1}$ we would have to consider $a_{1+n}\le a_{n}a_{1}$ which does not appear to yield anything useful. – Jason Born Oct 23 '16 at 10:50
  • You can prove it by induction, i.e. $a_{n+1} \leq a_na_1 \leq (a_1)^n a_1$ (with the second inequality using the induction hypothesis) – Dániel G. Oct 23 '16 at 13:55
  • But convergence follows from the monotone convergence theorem, no? – Jason Born Oct 23 '16 at 15:32
  • @user3482534 Yes –  Oct 23 '16 at 15:53
  • I don't know why you disapproved of accepting your answer then. I did not require a comprehensive answer, although maybe other users might appreciate it. – Jason Born Oct 23 '16 at 15:55
  • @user3482534 Because I have no idea how to conclude from here. I can't prove the monotony, maybe is not monotone in general –  Oct 23 '16 at 15:59