(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

Show that if [itex]a_n > 0[/itex] for all n,

[itex]\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}}[/itex]

2. Relevant equations

[itex]\liminf{a_n^{1/n}} \leq \limsup{a_n^{1/n}}[/itex]

[itex]\liminf{\frac{a_{n+1}}{a_n}} \leq \limsup{\frac{a_{n+1}}{a_n}}[/itex]

These are obvious, by properties of sup and inf.

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

[itex]\liminf{\frac{a_{n+1}}{a_n}} \leq \liminf{a_n^{1/n}}[/itex]

and

[itex]\limsup{a_n^{1/n}} \leq \limsup{\frac{a_{n+1}}{a_n}}[/itex]

I'm focusing right now on the limsup inequality.

3. The attempt at a solution

I took ln of each term,

[itex]\ln{\limsup{a_n^{1/n}}} = \limsup{\ln{a_n^{1/n}}} = \limsup{\frac{\ln{a_n}}{n}}[/itex]

and

[itex]\ln{\limsup{\frac{a_{n+1}}{a_n}}} = \limsup{\ln{\frac{a_{n+1}}{a_n}}} = \limsup{(\ln{a_{n+1}}-\ln{a_n})}[/itex]

This implies that I need to show that

[itex]\limsup{\frac{\ln{a_n}}{n}} \leq \limsup{(\ln{a_{n+1}}-\ln{a_n})}[/itex]

I've let

[itex]M_1 := \limsup{\frac{\ln{a_n}}{n}}[/itex]

[itex]M_2 := \limsup{(\ln{a_{n+1}}-\ln{a_n})}[/itex]

So now I just have to show that [itex]M_1 \leq M_2[/itex]

By the definition of limit, for all ε>0 there exists an N_{1}in the natural numbers such that for all [itex]n > N_1[/itex] we have

[itex]M_1 - ε < \frac{\ln{a_n}}{n} < M_1 + ε[/itex]

Similarly, for all ε>0 there exists an N_{2}such that for all [itex]n > N_1[/itex] we have

[itex]M_2 - ε < \ln{a_{n+1}}-\ln{a_n} < M_2 + ε[/itex]

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 N_{1}, N_{2}exist to satisfy the inequalities. However, I don't know how to go from here to show [itex]M_1 \leq M_2[/itex]. 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.

**Physics Forums | Science Articles, Homework Help, Discussion**

Dismiss Notice

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

Can you offer guidance or do you also need help?

**Physics Forums | Science Articles, Homework Help, Discussion**