inequality of series, showing their difference is less than a number given it is monotonically increasing Given $a_1=1$ $$a_{n+1} =
\frac{2+2a_n}{2+a_n}  \text{is  bounded: } 1\leq a_n \leq 2, \quad  a_{n+1}-a_n=\frac{2(a_n-a_{n-1})}{(2+a_n)(2+a_{n-1})}$$ 
Prove $$|a_{n+1}-a_n|\leq \frac{1}3\left(\frac{2}9\right)^{n-1}$$
So I can show that $$\lvert a_{n+1}-a_n \rvert \leq \frac{2\lvert a_n-a_{n-1}\rvert }9,$$ but I then end up with $2/9\ldots$  
Any hints? How should I be thinking of this? Like where does the $^{n-1}$ come from?

You have done most of the work already. Let $b_n=|a_{n+1}-a_n|$ for $n\in \mathbb{N} ^*$.
Then, you proved that $\forall n\in \mathbb{N}^*, b_n\leqslant \frac{2} {9} b_{n-1}$. You can prove $b_n \leqslant b_1 \left( \frac{2} {9} \right)^{n-1}$ using induction since $b_n$ is always positive.
Goal:
Prove that, $\forall n\in \mathbb{N}^*, b_n \leqslant b_1\left( \frac{2}{9} \right)^{n-1}$
Initialization:
At $n=1$, $b_1 = b_1 \times \left( \frac{2}{9} \right)^{1-1}$. Thus the property is initialized
Heredity:
If $b_{n-1} \leqslant b_1\left( \frac{2}{9} \right)^{n-2}$, then, since $b_n\leqslant \frac{2}{9} b_{n-1}$ then, $b_n \leqslant \frac{2}{9} \times b_1\left( \frac{2}{9} \right)^{n-2}=b_1\left( \frac{2}{9} \right)^{n-1}$
We thus proved the result for all $n>0$
As you noticed, $b_1=\frac{1}{3}$ hence the final result.
That is pretty much the proof for the general term of a geometric sequence. 
