# A limsup inequality (showing that the root test is stronger than the ratio test)

1. Oct 3, 2011

### moxy

1. The problem statement, all variables and given/known data

Show that if $a_n > 0$ for all n,

$\liminf{\frac{a_{n+1}}{a_n}} \leq \liminf{a_n^{1/n}} \leq \limsup{a_n^{1/n}} \leq \limsup{\frac{a_{n+1}}{a_n}}$

2. Relevant equations

$\liminf{a_n^{1/n}} \leq \limsup{a_n^{1/n}}$

$\liminf{\frac{a_{n+1}}{a_n}} \leq \limsup{\frac{a_{n+1}}{a_n}}$
These are obvious, by properties of sup and inf.

Which implies that, to complete the proof, I must show that

$\liminf{\frac{a_{n+1}}{a_n}} \leq \liminf{a_n^{1/n}}$
and
$\limsup{a_n^{1/n}} \leq \limsup{\frac{a_{n+1}}{a_n}}$

I'm focusing right now on the limsup inequality.

3. The attempt at a solution

I took ln of each term,

$\ln{\limsup{a_n^{1/n}}} = \limsup{\ln{a_n^{1/n}}} = \limsup{\frac{\ln{a_n}}{n}}$

and

$\ln{\limsup{\frac{a_{n+1}}{a_n}}} = \limsup{\ln{\frac{a_{n+1}}{a_n}}} = \limsup{(\ln{a_{n+1}}-\ln{a_n})}$

This implies that I need to show that

$\limsup{\frac{\ln{a_n}}{n}} \leq \limsup{(\ln{a_{n+1}}-\ln{a_n})}$

I've let

$M_1 := \limsup{\frac{\ln{a_n}}{n}}$
$M_2 := \limsup{(\ln{a_{n+1}}-\ln{a_n})}$

So now I just have to show that $M_1 \leq M_2$

By the definition of limit, for all ε>0 there exists an N1 in the natural numbers such that for all $n > N_1$ we have

$M_1 - ε < \frac{\ln{a_n}}{n} < M_1 + ε$

Similarly, for all ε>0 there exists an N2 such that for all $n > N_1$ we have

$M_2 - ε < \ln{a_{n+1}}-\ln{a_n} < M_2 + ε$

I think I can say that the epsilons are the same, since for each of these statements, I can choose a specific epsilon and I'm guaranteed that an N1, N2 exist to satisfy the inequalities. However, I don't know how to go from here to show $M_1 \leq M_2$. It might be trivial, I don't know, but I just don't see it. I've been working on this problem for too long, and it's making me cross-eyed. It was a homework problem that I turned in unfinished, and now I'm just trying to finally figure it out for my own sanity.