# divisor convergence?

• Sep 26th 2007, 08:17 AM
cowgirl123
divisor convergence?
If $a_n$ > 0 and (an+1/ $a_n)$ converges to p and p<1, prove $a_n$ converges to zero.
• Sep 26th 2007, 08:40 AM
Jhevon
Quote:

Originally Posted by cowgirl123
If $a_n$ > 0 and (an+1/ $a_n)$ converges to p and p<1, prove $a_n$ converges to zero.

consider this:

If $\frac {a_{n + 1}}{a_n} \to p < 1$, it must mean that $a_n >a_{n + 1}$. But this means that $\{ a_n \}$ is strictly decreasing. But $a_n > 0$ for all $n$. Which means that we are bounded below by 0. Thus $a_n$ is a monotonically decreasing sequence that is bounded below (by zero). Now say that more formally
• Sep 26th 2007, 10:17 AM
ThePerfectHacker
Quote:

Originally Posted by cowgirl123
If $a_n$ > 0 and (an+1/ $a_n)$ converges to p and p<1, prove $a_n$ converges to zero.

This means, $\lim \frac{a_{n+1}}{a_n} = p$. It must be the case that $p\geq 0$ because the sequence of ratios is a sequence of non-negative terms. So the limit is non-negative. I leave the case $p=0$ to you to prove. I will assume $p>0$, i.e. $0. The important step is to notice that there most exists $\epsilon > 0$ so that $p+\epsilon < 1$. This means for $\epsilon >0$ we have by convergence, $\left| \frac{a_{n+1}}{a_n} - p \right| < \epsilon$ for $n\geq N\in \mathbb{N}$. Thus, $\frac{a_{n+1}}{a_n} - p < \epsilon \implies a_{n+1} < (p+\epsilon)a_n$ for $n\geq N$.
This means,
$a_{N+1} < (p+\epsilon)a_N$
$a_{N+2} < (p+\epsilon)a_N < (p+\epsilon)^2 a_N$
$a_{N+3} < (p+\epsilon)a_{N+2} < (p+\epsilon)^2 a_{N+1} < (p+\epsilon )^3 a_N$.
In general,
$a_{N+k} < (p+\epsilon)^k a_N$ for $k\geq 1$.
This tells us that the sequence $\{ a_ n \}$ is bounded by $(p+\epsilon)^{n-N} a_N$ for all $n\geq N$.
Thus,
$0 < a_n < (p+\epsilon)^n a_n (p+\epsilon)^{-N}$.
Now, $\lim \ (p+\epsilon)^n a_n (P+\epsilon)^{-N} = 0$.
Because $0 < (p+\epsilon) < 1$ and it is a geometric sequence.
Q.E.D.