How to prove that $|w_{n+1}-w_n| \leq (2/3)^2 |w_n - w_{n-1}|$? If a sequence is given by $S_{n+1}=S_n + S_{n-1}$ where $S_1=1$ and $S_2=2$  and we let $a_n=S_n/S_{n-1}$ for all $n \geq 2$. For each $n \geq 2$, prove that $|a_{n+1}-a_n| \leq (2/3)^2 |a_n - a_{n-1}|$?
This is what I'm thinking...
Let n=3. Then $|a_{n+1}-a_n| \leq  (2/3)^{2}|a_n - a_{n-1}|$ is true for $n=3$ since 
$|a_{4}-a_3| \leq  (2/3)^{2}|a_3 - a_2|$
$|\frac{5}{3}-\frac{3}{2}| \leq  (2/3)^{2}|\frac{3}{2}-2|$
$\frac{1}{6} \leq  (2/3)^{2}(1/2$
$\frac{1}{6} \leq  2/9$
Now let $n=k$ where $k \in \mathbb{R}$ such that $ k \geq 2$. Then
$|a_{k+1}-a_k| \leq  (2/3)^{2}|a_k - a_{k-1}|$
$|\frac{S_{k+1}}{S_k}-\frac{S_k}{S_{k-1}}| \leq  (2/3)^{2}|\frac{S_{k}}{S_{k-1}}-\frac{S_{k-1}}{S_{k-2}}|$
$\frac{1}{S_kS_{k-1}} \leq  (2/3)^{2}\frac{S_{k}S_{k-2}-S^2_{k-1}}{S_{k-1}S_{k-2}}$
Thus, it holds for $n=k+1$
 A: Hint: First observe that the sequence $S_n$ is the Fibonacci sequence, if you simply let $S_0=1$. It satisfies numerous wonderful properties, but the one that you need to use is $S_{n+1}S_{n-1}-S_n^2=(-1)^n$. After substituting this, you will be left to prove: $$\dfrac{4}{9}S_{n}\geq S_{n-2}$$, for $n\geq 3.$
This can be proved by many different ways, the simplest of which will probably be induction. 
A: Like user dezdichado already mentioned, this is the fibonacci sequence, but we can approach this independently without necessarily using known facts about the sequence. 
You also have a typo on your final inequality which it should instead read as: 

$$\vert \frac{S_{k+1}S_{k-1}-S_k^2}{S_kS_{k-1}} \vert \leq (2/3)^2\vert  \frac{S_{k}S_{k-2}-S_{k-1}^2}{S_{k-1}S_{k-2}} \vert$$

From there we can use the following simple fact
$S_{k+1}S_{k-1}-S_k^2 = (S_{k}+S_{k-1})S_{k-1}-(S_{k-1}+S_{k-2})^2 = S_kS_{k-1} - 2S_{k-1}S_{k-2} - S_{k-2}^2 =  $
$S_{k-1}(S_k - S_{k-2}) - S_{k-2}(S_{k-1} + S_{k-2}) = S_{k-1}^2 - S_kS_{k-2}$
We conclude that: 

$$\vert S_{k+1}S_{k-1}-S_k^2 \vert = \vert S_{k-1}^2 - S_kS_{k-2} \vert = \vert S_kS_{k-2} - S_{k-1}^2 \vert$$

and thus your original simplifies to 

$$\vert \frac{1}{S_k} \vert \leq (2/3)^2\vert  \frac{1}{S_{k-2}} \vert \Rightarrow  \frac{1}{S_k} \leq (2/3)^2 \frac{1}{S_{k-2}} \Rightarrow  \frac{S_{k-2}}{S_k} \leq (2/3)^2 \Rightarrow  \frac{S_{k}}{S_{k-2}} \geq (3/2)^2 = \frac{9}{4}$$

Doing a bit more work we get: 
$\frac{S_{k}}{S_{k-2}} = \frac{S_{k-1}+S_{k-2}}{S_{k-2}} = 1 + \frac{S_{k-1}}{S_{k-2}}$. So it's enough to show that: 
$$\frac{S_{k-1}}{S_{k-2}} \geq \frac{5}{4}$$
The last should be easy to verify using a simple inductive argument
