The sequence $x_{n+1}=ax_{n}+b $ converges to where? $$a,b \in \mathbb R , \ 0\lt a\lt 1 . $$ Define  the  sequence  $$x_{n+1}=ax_{n}+b \text{ for } n\ge0\ .$$
Then  for  a  given  $\ \ x_0\ \ $ , does  this  sequence  converge? And  if  it  does, to  where?
Now  what  I  did  was  write  down  the  first  few  terms , say, $x_0,x_1,x_2,\ldots$ and  found  out  that  the  sequence  goes  by  the  rule  $$x^n=a^nx_{0}+\sum_{i=0}^{n-1} a^i b$$  Now  I  suppose  the  sequence  $x_{0}\{a^n\}$  will  converge  as  $0< a < 1$. And  the  series  is  $b\left(\sum_{i=0}^{n-1} a^i \right)$. Not  sure  about  the  series.  And  what  will  be  the  limit?
Thanks  for  any  help.
 A: You are extremely close to the answer, you just have to recall that for any $a\in (0,1) $ we have:
$$ \sum_{i\geq 0}a^i = \lim_{n\to +\infty}\sum_{i=0}^{n-1}a^i = \frac{1}{1-a}.$$
Given that, you may also notice that:
$$ \left(x_{n+1}-\frac{b}{1-a}\right) = a\cdot\left(x_{n}-\frac{b}{1-a}\right)$$
holds, so you know how fast the convergence is.
A: This only answers the "to what does this converge?" part of the question and does not prove the sequence actually converges.
A neat trick for these sorts of problems is to assume the sequence converges and then use simple algebra to find the limit.
If $x_n$ converges, then $\lim_{n\to\infty} x_n = \lim_{n\to\infty}x_{n+1} = x$.  Then, we just solve the equation: $x=ax+b$ for $x$:
$$x = \frac{b}{1-a}$$
A: As you showed the general formula for the $n$th term, the sum in the bracket is a geometric series and will also converge provided $|a| < 1$.
Oussama Boussif's answer then gives the correct limit to $\frac b {(1-a)}$. Interestingly independent of $x_0$. Sorry, I can't comment due to low reputation. Here is a good link for some standard formulae: https://en.wikipedia.org/wiki/List_of_mathematical_series.
A: Yes that's the limit. You can easily show that the expression is correct by induction. 
The series is a geometric series, which has a well known formula for the sum to infinity. 
The limit of $a^n$ as $n\rightarrow\infty$ is 0.
