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 $$\lim(x_{n+1}/x_n)<1$$. Show for some $$0<r<1$$, and for some $$C>0$$, $$0<x_n<Cr^n$$

Homework Equations

The Attempt at a Solution

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

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

$$0<\frac{x_{n+1}}{x_n}<r<1.$$

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:

$$\frac{x_{n+k}}{x_n} = \frac{x_{n+1}}{x_n} \frac{x_{n+2}}{x_{n+1}} \cdots \frac{x_{n+k}}{x_{n+k-1}} < r^k$$

 

[blinktx411]

Can I let $$x_n=C$$ and therefore say the sequence $$x_{n+k}<Cr^k$$? 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 $$x_{n+k}<Cr^k$$? 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

$$x_{n} < C r^n$$

not

$$x_{n+k} < Cr^k$$

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 $n \geq N$,

$$0 < \frac{x_{n+1}}{x_n} < r < 1$$

which yields

$$x_{N+k} < r^k x_N$$

for all $$k \geq 0$$.

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

$$x_{N+k} < r^{N+k} (x_N r^{-N})$$

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

$$x_{N+k} < C_1 r^{N+k}$$

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

$$x_n < C_1 r^n$$

for all $n < N$.

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

 

[blinktx411]

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

 

[jbunniii]

So, for every $$j<N$$, Let $$M_j$$ be such that $$x_j<M_jr^j$$ (there exists such an M by the archimedean property). Now, if I take $$C=\sup\{M_j,C_1| j\in\mathbb{N},j<N\}$$, 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.