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?

Question

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.

Answer 1

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.

Answer 2

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.