Let the sequence be defined recursively $X_{n+1}=pX_{n}+q$ for p nonzero and $X_{1}$ arbitrary. In what conditions does the sequence converge? Let the sequence be defined recursively $X_{n+1}=pX_{n}+q$ for p nonzero and $X_{1}$ arbitrary. In what conditions does the sequence converge?
I bumped into this problem and I don't know where to even begin. I would really appreciate some help.
 A: What you are looking at is an arithmetico-geometric sequence.
When it converges (notably for $|p|<1$), it converges to the invariant point $\frac{q}{1-p}$.
Actually, if $p=1$, it is just another arithmetic sequence. If $p\neq 1$, the general term is $X_n = p^n(X_0-\frac{q}{1-p}) + \frac{q}{1-p}$ (because $X_n-\frac{q}{1-p}$ is a geometric sequence).
Funnily enough, the limit does not depend on the initial value of the sequence.
A: From the characteristic polynomials (here is an example to avoid it looking dry) perspective
$$X_{n+1}=pX_{n}+q$$
$$X_{n+2}=pX_{n+1}+q$$
thus
$$X_{n+1}-pX_{n}=X_{n+2}-pX_{n+1} \iff X_{n+2}-(p+1)X_{n+1}+pX_{n}=0$$
with characteristic polynomial
$$x^{2}-(p+1)x+p=0$$
with solutions $x_1=1$ and $x_2=p$
and the general term of the sequence
$$X_n=Ax_1^n+Bx_2^n=A+Bp^n$$
We can find $A,B$ from
$$\left\{\begin{matrix}
X_1=A+Bp\\ 
pX_1+q=A+Bp^2
\end{matrix}\right.$$
leading to $B=\frac{X_1}{p}+\frac{q}{p(p-1)}$, $A=-\frac{q}{p-1}$ and
$$X_n=-\frac{q}{p-1}+\left(X_1+\frac{q}{p-1}\right)p^{n-1} \tag{*}$$
As a result 


*

*if $|p|>1$ the sequence will diverge 

*if $B=0 \iff X_1=-\frac{q}{p-1}$, the sequence is constant.

*if $p=1 \Rightarrow X_n=X_1+(n-1)q$ diverges.

*if $|p|<1$ - converges. 

