1
$\begingroup$

Let $\alpha > 0$ and $a_n$ be a sequence of positive real numbers such that $a_{n+2} \leq a_{n+1} - \alpha a_n$ for all $n\geq 0$. Is it true that $a_n \leq A_1x_1^n + A_2x_2^n$ for some constants $A_1$ and $A_2$, where $x_1$ and $x_2$ are the roots of $t^2 - t + \alpha = 0$?

If $\leq$ was instead $=$, then this is well known from the theory of linear recurrences (and follows by induction). However, here assuming $a_n \leq A_1x_1^n + A_2x_2^n$, we cannot say $a_{n+1} -\alpha a_n \leq A_1x_1^{n+1} + A_2x_2^{n+1} - \alpha(A_1x_1^{n} + A_2x_2^{n})$ when $\alpha > 0$ because of the minus sign. Is this fixable at all?

Any help appreciated!

$\endgroup$

1 Answer 1

3
$\begingroup$

Well, with the hypotheses you presented you can show that the sequence converges to $0$. If this counts as 'fixable' then yes it is fixable. If you want to get precise bounds compared to the solution to linear equations then that is more difficult but still possible.

{$a_{n}$} is positive monotone non-increasing and bounded below by 0, hence converges to $L\geq 0$. If $L > 0$ you can use epsilon delta arguments with the limsup and liminf to show:

$$L < L -\alpha L$$.

Which will lead to a contradiction, forcing $L=0$.

$\endgroup$
2
  • $\begingroup$ How do you show that it converges to 0? That's kinda enough for what I am aiming at (not presented in this question). $\endgroup$ Jan 31, 2021 at 20:53
  • 2
    $\begingroup$ Some stronger results can be found in this article: msp.org/involve/2008/1-1/p07.xhtml $\endgroup$ Jan 31, 2021 at 21:01

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .