Proof of Sequence Convergence: 0<r<1 and 0<x_n<Cr^n?

[blinktx411]

Homework Statement

Let X=(x_n) be a sequence of strictly positive numbers such that \lim(x_{n+1}/x_n)&lt;1. Show for some 0&lt;r&lt;1, and for some C&gt;0, 0&lt;x_n&lt;Cr^n

Homework Equations

The Attempt at a Solution

Let \lim(x_{n+1}/x_n)=x&lt;1
By definition of the limit, \lim(x_{n+1}/x_n)=x \Rightarrow \forall \epsilon&gt;0 there exists \: K(\epsilon) such that . \: \forall n&gt;K(\epsilon)

|\frac{x_{n+1}}{x_n}-x|&lt;\epsilon
Since i can pick any epsilon, let epsilon be such that \epsilon + x = r &lt;1. Also, I know that since this is a positive sequence, \frac{x_{n+1}}{x_n}&gt;0. Therefore, for large enough n,

0&lt;\frac{x_{n+1}}{x_n}&lt;r&lt;1.

From here I am not sure where to go, any hints would be much appreciated! I cannot find out what this tells me about $x_n$

 

[jbunniii]

For large enough n you also have the following:

\frac{x_{n+k}}{x_n} = \frac{x_{n+1}}{x_n} \frac{x_{n+2}}{x_{n+1}} \cdots \frac{x_{n+k}}{x_{n+k-1}} &lt; r^k

 

[blinktx411]

Can I let x_n=C and therefore say the sequence x_{n+k}&lt;Cr^k? If so then n would fixed and k would be the index, correct? Thanks for the speedy response!

 

[jbunniii]

Can I let x_n=C and therefore say the sequence x_{n+k}&lt;Cr^k? If so then n would fixed and k would be the index, correct? Thanks for the speedy response!

You're not quite there yet.

First note that you need

x_{n} &lt; C r^n

not

x_{n+k} &lt; Cr^k

Also, it needs to be true for EVERY n, not just sufficiently large n.

First let's introduce some notation. It is easier to discuss if we give this "sufficiently large" n a name, say N. For all n \geq N,

0 &lt; \frac{x_{n+1}}{x_n} &lt; r &lt; 1

which yields

x_{N+k} &lt; r^k x_N

for all k \geq 0.

I can get the index and exponent to agree by writing, equivalently,

x_{N+k} &lt; r^{N+k} (x_N r^{-N})

Then if I set C_1 = x_N r^{-N} I get

x_{N+k} &lt; C_1 r^{N+k}

which looks pretty promising. However, this C_1 may not be large enough to work for ALL n, i.e. it may not be true that

x_n &lt; C_1 r^n

for all n &lt; N.

But note that there are only finitely many x_n with n &lt; N. Can you use that fact to find a C that does work for all n?

 

[blinktx411]

So, for every j&lt;N, Let M_j be such that x_j&lt;M_jr^j (there exists such an M by the archimedean property). Now, if I take C=\sup\{M_j,C_1| j\in\mathbb{N},j&lt;N\}, that should do the job since there are only finitely many M's, correct?

 

[jbunniii]

So, for every j&lt;N, Let M_j be such that x_j&lt;M_jr^j (there exists such an M by the archimedean property). Now, if I take C=\sup\{M_j,C_1| j\in\mathbb{N},j&lt;N\}, that should do the job since there are only finitely many M's, correct?

Looks good to me. You can even call it "max" instead of "sup" since there are only N+1 numbers under consideration.