Finding the limit of the recursive sequence defined as $5a_{n+2}=3a_{n+1}+2a_n$ 
Find the limit of the recursive sequence defined as follows: $$5a_{n+2}=3a_{n+1}+2a_n\;\;\text{for $n\geq 0$, $a_1=1/2$  and $a_2=2/3$}.$$

Could someone give me a hint as to how to start this problem? I think it is increasing.

$$5a_{n+2}-5a_{n+1}=2a_{n}-2a_{n+1}$$ or
$$a_{n+2}-a_{n+1}=-\frac{2}{5}(a_{n+1}-a_n),$$
which says that $b_n=a_{n+1}-a_n$ is a geometric progression. 
Thus, $$a_{n+1}-a_n=(a_2-a_1)\left(-\frac{2}{5}\right)^{n-1}=\frac{1}{6}\left(-\frac{2}{5}\right)^{n-1}.$$
Thus, $$a_n=a_1+\sum_{k=1}^{n-1}(a_{k+1}-a_k)=\frac{1}{2}+\frac{1}{6}\sum_{k=1}^{n-1}\left(-\frac{2}{5}\right)^{k-1}=$$
$$=\frac{1}{2}+\frac{1}{6}\cdot\frac{\left(-\frac{2}{5}\right)^{n-1}-1}{-\frac{2}{5}-1}\rightarrow\frac{1}{2}+\frac{1}{6}\cdot\frac{-1}{-\frac{2}{5}-1}=\frac{13}{21}.$$
