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[Reposting this because my previous write up had typos and errors (it was late)]

I am trying to prove that that $\lim \inf \frac{u_{n}}{u_{n-1}} \le \lim \inf \sqrt[n]{u_n}$ where $u_n > 0$. I have the basic idea but am having difficulty making the proof tight. My answer so far is this.

Let $\lim \inf \sqrt[n]{u_n} = k < \infty$. I want to show that for all $\varepsilon > 0$ and for all $N = 1, 2, ...$ there exists $n \ge N$ such that $${u_n}/{u_{n-1}} < k + \varepsilon.$$ Thus $\lim \inf \frac{u_n}{u_{n-1}} \le k = \lim\inf\sqrt[n]{u_n}$.

Toward this end, I know by definition of the "$\lim\inf$" that for any $\varepsilon'$ and $N$ there must exist $n \ge N$ such that

  1. $\sqrt[n+1]{u_{n+1}} < k + \varepsilon'$, and
  2. $\sqrt[n]{u_n} > k - \varepsilon'$

(If we had #1 but not #2 then $k$ wouldn't be greater than the lim inf.)

Thus $$\frac{u_{n+1}}{u_n} < \frac{ (k+\varepsilon')^{n+1} }{ (k-\varepsilon')^{n} } = \left(\frac{ k+\varepsilon' }{ k-\varepsilon' } \right)^n (k+\varepsilon') < k + \varepsilon. $$

However, I am having trouble picking $\varepsilon'$ so that the last inequality holds for $\varepsilon$. Any help is appreciated.

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    Well, for limsup we have the opposite inequality: $\limsup \sqrt[n]{u_n} \leq \limsup \frac{u_{n+1}}{u_n}$. – 2017-01-10
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    Okay, but how do we prove that? The same issue comes up. – 2017-01-10
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    It is easier to utilize $k = \liminf (u_{n+1}/u_n)$ and give a lower bound of $\sqrt[n]{u_n}$. If you have heard of the [*Cesaro-Stolz theorem*](https://en.wikipedia.org/wiki/Stolz%E2%80%93Ces%C3%A0ro_theorem#The_General_Form), you can also realize that this is just a disguise of this theorem. – 2017-01-10
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    There is an answer [here](http://math.stackexchange.com/questions/1764134/theorem-3-37-in-baby-rudin-lim-inf-fracc-n1c-n-leq-lim-inf-sqrtnc-n?rq=1). – 2017-01-10
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    See also [Inequality involving $\limsup$ and $\liminf$: $ \liminf(a_{n+1}/a_n) \le \liminf((a_n)^{(1/n)}) \le \limsup((a_n)^{(1/n)}) \le \limsup(a_{n+1}/a_n)$](http://math.stackexchange.com/q/69386) and other posts [linked there](http://math.stackexchange.com/questions/linked/69386). – 2017-01-10

1 Answers 1

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Let $L=\lim \inf u_n^{1/n}.$

(1). If $0L.$ Then take $d>0,$ and $n_0,$ such that $$n\geq n_0\implies u_{n+1}/u_n>(1+d)L. $$ Then for $m>0$ we have $$u_{n_0+m}=u_{n_0} \prod_{j=0}^{m-1}\frac {u_{n_0+j+1}}{u_{n_0+j}}>u_{n_0}((1+d)L)^m$$ which implies $$(u_{n_0+m})^{1/(n_0+m)}>A_mL(1+d)B_m$$ where $A_m=u_{n_0}^{1/(n_0+m)}$ and $B_m=(L(1+d))^{-n_0/(n_0+m)}.$

Now $\lim_{m\to \infty}A_m=\lim_{m\to \infty}B_m=1.$ So, letting $m\to \infty$, we have $L=\lim \inf u_m=\lim \inf u_{m+n_0}\geq L(1+d)>L$ which is a contradiction.

(2) . If $L=0.$ Suppose by contradiction that $\lim_{m\to \infty} \inf u_{n+1}/u_n=K>0.$ We can take $n_0$ such that $n\geq n_0\implies u_{n+1}/u_n>K/2.$ By the method in (1) we obtain $0=\lim \inf u_n>K/2>0,$ which is a contradiction.

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    Thanks, I'm looking for a direct proof of the same along the lines I outlined above but this is nice to see as well. – 2017-01-12