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

[moxy]

Homework Statement

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

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

Homework 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.

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.

 

[mighty2000]

One way to approach this problem is to use the fact that the logarithm function is strictly increasing. This means that if a_n > 0 for all n, then

\frac{\ln{a_n}}{n} < \frac{\ln{a_{n+1}}}{n+1}

for all n. This can be seen by rearranging the terms and noting that n+1 > n.

Using this fact, we can rewrite the inequality as

M_1 - ε < \frac{\ln{a_n}}{n} < \frac{\ln{a_{n+1}}}{n+1} < M_2 + ε

for all n > N1.

Now, taking the limit as n -> ∞, we have

M_1 \leq \limsup{\frac{\ln{a_n}}{n}} \leq \limsup{\frac{\ln{a_{n+1}}}{n+1}} \leq M_2

since \limsup is monotone. This implies that M_1 \leq M_2, as desired.

Alternatively, we can also use the fact that the sequence a_n is positive and strictly increasing to show that \limsup{\frac{a_{n+1}}{a_n}} = \limsup{a_n^{1/n}}. This allows us to rewrite the inequality as

M_1 - ε < \frac{a_n}{a_{n+1}} < \frac{a_n^{1/n}}{a_{n+1}^{1/(n+1)}} < M_2 + ε

for all n > N1.

Taking the limit as n -> ∞, we have

M_1 \leq \limsup{\frac{a_n}{a_{n+1}}} = \limsup{a_n^{1/n}} \leq \limsup{\frac{a_n^{1/n}}{a_{n+1}^{1/(n+1)}}} \leq M_2

since \limsup is monotone. This implies that M_1 \leq M_2, as desired.

Overall, the key idea is to use the monotonicity of the limit and the properties of the logarithm function to show that the two limits are equal, which then allows us to use the properties of \limsup and \liminf to