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