Real analysis: inequality limitsuperior/inferior

[K29]

Homework Statement

Consider $\sum_{1}^{\infty} a_{n}, a_{n} \neq 0$

Show that

$\underline{\lim\limits_{n \rightarrow \infty}}|\frac{a_{n+1}}{a_{n}}| \leq \underline{\lim\limits_{n \rightarrow \infty}}\sqrt[n]{|a_{n}|}\leq \overline{\lim\limits_{n \rightarrow \infty}}\sqrt[n]{|a_{n}|)} \leq \overline{\lim\limits_{n \rightarrow \infty}}|\frac{a_{n+1}}{a_{n}}|$

The Attempt at a Solution

I have just been thinking about the limit superior part of the inequality:

$\overline{\lim\limits_{n \rightarrow \infty}}\sqrt[n]{|a_{n}|)} \leq \overline{\lim\limits_{n \rightarrow \infty}}|\frac{a_{n+1}}{a_{n}}|$

Ok so I know that for all n
$|\frac{a_{n+1}}{a_{n}}| \leq M$
and that for all n
$\sqrt[n]{|a_{n}|} \leq N$

The best way I can see to prove that $N \leq M$ is to do a proof by contradiction
so if N>M then for a sufficiently large p such that:
$|\frac{a_{p+1}}{a_{p}}| \leq M \leq \sqrt[p]{|a_{p}|} \leq N$
I've spent a lot of time manipulating this equation, but I'm not getting to a contradiction.

Can anyone see a way to prove this or an alternative way to do the proof that works?

Another thing I tried was saying:
$\sqrt[n]{|a_{n}|} \leq N$
So
$a_{n} \leq N^{n}$
$\sqrt[n+1]{|a_{n+1}|} \leq N$
So
$a_{n+1} \leq N^{n+1}$

But I can't really do anything with that as far as I can see. Was just a thought.

 

[micromass]

Hi K29!

Let's get you started, will we.

First, we will change that annoying limsup by a limit. Note in general that

$\limsup{x_n}\leq \limsup{y_n}$

If for every subsequence $x_(n_k)$ that increases to the limsup, we have that

$$\lim{x_{n_k}}\leq \limsup{y_n}$$

In general, we can take the $x_n$ to converge (maybe to infinity) and we can take them to be increasing. So we need to show that

$$\lim{\sqrt[n]{|a_n|}}\leq \limsup{\frac{|a_{n+1}|}{|a_n|}}$$

Now, what if you take the logarithm of both sides?

 

[K29]

Thanks for the help! Solved :)