# A recurrence relation

Motivated by a specific example, I have a rather general question to ask: suppose $a_n$ is a sequence defined by the relation $a_{n+1}=f_na_n+g_na_{n-1}$, $a_0=a>0$, $a_1=b>0$, where both $f_n$, $g_n$ are positive and $\lim\limits_{n \rightarrow \infty}f_n =0$, $\lim\limits_{n \rightarrow \infty}g_n =1$.

Under what additional conditions for the initial values and/or the functions $f_n$, $g_n$ involved (if any) can we deduce that the sequence $a_n$ is bounded?

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If you divide by $g_n$, then you'll see that we may assume that $g_n = 1$. In this case, let $F_n = f_nf_{n-1} + f_{n-2}$ for $n > 2$. Then what you want is for $\{F_n\}$ to be bounded. In fact, by positivity of all the terms in your original sequence, this condition is equivalent to $\{a_n\}$ being bounded. –  William May 21 '12 at 0:42
So you're basically saying it holds unconditionally? –  the_fox May 21 '12 at 4:09
I don't think William is saying that. If $a=b=1$, $f_n=1/n$, $g_n=1$, I think it's not hard to show $a_n$ is unbounded. –  Gerry Myerson May 21 '12 at 4:59
Sure. But if ${a_n}$ is bounded if and only if ${F_n}$ is as William claims, then the sequence $1/n(n-1)+1/(n-2)$ would have to be unbounded too, no? –  the_fox May 21 '12 at 6:49
Also, I suspect the behavior of $a_n$ might be related to $\sum\limits_{n=1}^{\infty}f_n$. –  the_fox May 21 '12 at 6:53