If $x_{n+1} = ax_n+b$ for $n \geq 0$, then prove that the sequence $\{x_n\}$ converges to $\frac{b}{1-a}$ for a given value of $x_0$.

Question

Here $a,b \in \mathbb{R}$ and $0 \lt a \lt 1$.

First thing which came to my mind was that the sequence could be decreasing because of the condition $0 \lt a \lt 1$, but there is no restriction on $b$, so this need not be the case. Also because there is no restriction on the value of $x_0$, I don't know how the sequence will look like for higher order terms.

Edit: As pointed out by Arpit, the limit itself was not hard to find. I was having difficulty proving that the sequence will converge for all values of $b$ and $x_0$.

Answer 1

Note that the only part which need some justification is to show that the sequence $(x_n)$ Converges.Once you know $(x_n)$ converges then assuming lim$_{ n\to \infty}x_n=L$ and using $x_{n+1}=ax_n+b$ gives $L=aL+b$ and hence $L=\frac {b}{1-a}$

Claim: $(x_n)$ is convergent.

Below i will write a exercise which will take care of convergence part.

Exercise: Let $(x_n)$ be a sequence such that $\vert x_n-x_{n-1} \vert \leq a_n$ where $(a_n)$ is such that $\sum a_n < \infty$ then $(x_n)$ is convergent.

Hint for solution: Show that $(x_n)$ is cauchy and hence convergent as $\mathbb R$ is complete.

Answer 2

We assume $0<a<1$. One may set $$ y_n=x_n-\frac{b}{1-a},\qquad n=1,2,\cdots, $$ then one may prove that $\{y_n\}$ is just a geometric sequence: $$ y_{n+1}=a\cdot y_n,\qquad n=1,2,\cdots. $$ From there one gets $$ y_n=\left(x_0-\frac{b}{1-a} \right)a^n,\qquad n=1,2,\cdots $$ deducing a closed form of $x_n$ then answering the OP question.