The sequence $x_{n+1}=ax_{n}+b $ converges to where?

Question

$$a,b \in \mathbb R , \ 0\lt a\lt 1 . $$ 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.

Answer 1

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 $x$: $$x = \frac{b}{1-a}$$

Answer 2

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.

Answer 3

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.

Answer 4

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.