Simple linear recursion $x_n=\frac{x_{n-1}}{a}+\frac{b}{a}$ with $a>1, b>0$ and $x_0>0$
I tried to solve it using the generating function but it does not work because of $\frac{b}{a}$, so may you have an idea.

Hint: Let $x_n=y_n+c$, where we will choose $c$ later. Then
$$y_n+c=\frac{y_{n-1}+c}{a}+\frac{b}{a}.$$
Now can you choose $c$ so that the recurrence for the $y$'s has no pesky constant term?
Remark: There is a fancier version of the above trick. Our recurrence (if $b\ne 0$) is not homogeneous. To solve it, we find the general solution of the homogeneous recurrence obtained by removing the $b/a$ term, and add to it some fixed particular solution of the non-homogeneous recurrence. In this case it is easy to find such a particular solution. Look for a constant solution.  
