Find a limit of recursion sequence

Question

I have to find a limit of recursion sequence $a_{n}=-\frac{3}{8}(a_{n-1}+a_{n-2})$ where $a_1=1, a_0=0$. I wrote few of first terms of sequence: $0,1,-\frac38,\frac{15}{64},-\frac{57}{8^3}...$ My idea was to prove that $|a_n|\leq(\frac12)^{n-1}$ what would mean that the limit is 0. I was trying to do that by induction but I can't do it. Everything was falling apart at second step (I tried to use triangle inequality but it's not good enough). Thank you in advance for any help.

Answer 1

Claim: $|a_i|\leq (0.9)^{i}$ for $i \geq 2$.

$a_2=-\frac 3 8 (1+0)$, so $|a_2| \leq (0.9)^{2}$.

Supppose $|a_i|\leq (0.9)^{i}$ for $2 \leq i <n$. Then, $|a_n|\le \frac 3 8 ((0.9)^{n-1}+(0.9)^{n-2})=\frac 3 8 (0.9)^{n-2}(0.9+1) $. Note that $\frac 38 (1.9)<(0.9)^{2}$. Hence, $|a_n|\le(0.9)^{n}$.