The convergence of sequence $a_n$ such that $pa_n + qa_{n + 1}$ converges This is a problem from Courant and John's volume 1, chapter 1. 
Let $a_n$ be a given sequence such that the sequence $b_n = pa_n + qa_{n + 1}$ where $|p| < q$, is convergent. Prove that $a_n$ converges. If $|p| > q > 0$, show that $a_n$ need not converge.
 A: Since this is homework, I only provide  a sketch of the answer.
When $|p|>q$, it should not be too hard for you to find a counterexample, by considering the case $b_n=0$. 
Suppose $|p|<q$. The sequence $w_n=-\big(\frac{-p}{q}\big)^{n-1}a_1$ then tends to zero, and the sequence $a^{\sharp}_n=a_n-w_n$ satsifies $b_n=pa^{\sharp}_n+qa^{\sharp}_{n+1}$ just like $a_n$ does. So we may assume without loss that $a_1=0$. 
We easily have by induction that for any $n\geq 1$,
$$
a_n=\bigg(\frac{-p}{q}\bigg)^n \Bigg(\sum_{j=1}^{n-1} p\bigg(\frac{q}{-p}\bigg)^j b_j \Bigg) 
\tag{1}
$$
It follows from (1) (by induction, again) that for any $n,t\geq 0$
$$
a_{n+t}-a_n=\frac{-1}{p}\bigg(\frac{-p}{q}\bigg)^n \Bigg(
\sum_{l=1}^{t} \bigg(\frac{-p}{q}\bigg)^{t-l} b_l 
+
\sum_{j=1}^{n-1} \bigg(\frac{q}{-p}\bigg)^j (b_{j+t}-b_j) 
\Bigg) 
\tag{2}
$$ 
Since $(b_n)$ is convergent, there is an absolute constant $M$
such that $|b_n| \leq M$ for all $n$. Let $\varepsilon >0$. Since
$(b_n)$ is convergent, it satisfies Cauchy’s criterion, so there is
and $N$ such that $|b_{j+t}-b_j| \leq \varepsilon$ for
any $j\neq N, t\geq 0$. Combining this with (2), we see that
for any $n > N_\varepsilon, t\geq 0$ we have
$$
|a_{n+t}-a_n| \leq \frac{1}{|p|}\bigg(\frac{|p|}{q}\bigg)^n \Bigg(
\sum_{l=1}^{t} \bigg(\frac{|p|}{q}\bigg)^{t-l} M 
+
\sum_{j=1}^{N-1} \bigg(\frac{q}{|p|}\bigg)^j (2M) 
+
\sum_{j=N}^{n-1} \bigg(\frac{q}{|p|}\bigg)^j \varepsilon 
\Bigg) 
\tag{3}
$$
It follows from (3) (and a little more work) that
$(a_n)$ satisfies Cauchy’s criterion and is therefore convergent.
