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For $x_n>0$ for all $n\in\Bbb N$ prove that $\underline\lim\frac{x_{n+1}}{x_n}\le\underline\lim\sqrt[n]{x_n}\le\overline\lim \sqrt[n]{x_n}\le \overline\lim\frac{x_{n+1}}{x_n}$

HINT: if $q<\underline\lim\frac{x_{n+1}}{x_n}$ then $q\le\frac{x_{n+1}}{x_n}$ for all $n\ge N_q$.

  • Here $\underline\lim$ mean limit inferior and $\overline\lim$ mean limit superior. Observe that the statement dont require that the sequence $(x_n)$ will be convergent.

I dont know how to prove this, some hint will be appreciated. My work so far: we can choose some $q>0$ that for some $N\in\Bbb N$ we have that

$$\left(q\le \frac{x_{n+1}}{x_n}\right)\land (q\le \sqrt[n]{x_n}),\quad\forall n\ge N_q$$

I dont get something useful from here (or at least I dont see that this is useful), an example

$$q^n\le x_n\le\frac{x_{n+1}}{q},\quad\forall n\ge N_q$$

I tried to get something by contradiction, i.e. that the statement $$\underline\lim\sqrt[n]{x_n}<\underline\lim\frac{x_{n+1}}{x_n}$$

is not possible. But again I dont found something enlightening. The unique thing that I worked is that if $(x_n)\to a$ is a constant sequence then the statement of the question is true.

Another crazy idea that I tried is choose some $0<q<1$, such that

$$\left(q\le\frac{x_{n+1}}{x_n}\right)\land (q\le x_n),\quad\forall n\ge N_q$$

then with little work we can see that

$$q\le\frac{x_{n+1}}{x_n}\implies q<\sqrt[m+n]{q^{m+1}}\le\sqrt[n+m]{x_{n+m}},\quad\forall n\ge N_q$$

but it is not clear that this last statement prove something useful. Well, any hint will be appreciated, thank you.

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Masacroso
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1 Answers1

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From the hint you have $ q \le x_{n+1} / x_n $ for each $ n \ge N_q $. Now you could use the telescopic product $ (x_{n+1}/x_n)...(x_{N_q+1}/x_{N_q})=x_{n+1}/x_{N_q} $.

Lázaro Albuquerque
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  • I did this, but now what I can do with this? I get $q^m x_{N_q}=x_{n+1}$. Taking the m-root now dont seem to prove anything (supposing that $m=n+1$). – Masacroso Oct 22 '16 at 19:40
  • Actually you should get $q^{n-N_{q}+1} \le x_{n+1} / x_{N_q}$, so that after taking (n+1)-root both sides you obtain $q \le x_{n+1}^{1/(n+1)} \cdot C^{1/(n+1)}$, with $C > 0$. – Lázaro Albuquerque Oct 22 '16 at 19:56
  • I know tetsuzan but still I dont see how this is useful. The $C>0$ can be less or greater to $1$, I know that $\sqrt[n] a\to 1$ for fixed $a$ but from here I dont see how to prove the stated inequality in the problem. This is why I did the last approach that I showed in my question. – Masacroso Oct 22 '16 at 20:00
  • So, taking $lim\ inf$ both sides you get almost the desired inequality, but with $q$ instead of $lim\ inf\ x_{n+1}/x_n$. However, since this is valid for an arbitrary $q < lim\ inf\ x_{n+1}/x_n$, it is valid for the supremum, $lim\ inf\ x_{n+1}/x_n$. – Lázaro Albuquerque Oct 22 '16 at 20:10
  • No we don't because to prove $a \le b$ it is enough to prove that $\forall \epsilon > 0,\ a - \epsilon < b$. – Lázaro Albuquerque Oct 22 '16 at 20:23
  • Oh, I see now. Thank you tetsuzan for your patience. – Masacroso Oct 22 '16 at 21:19