Convergence of two nested geometric sequences Let $\nu_t = b^t \nu_0$ be a geometric sequence where $\nu_0>0$, $0<b<1$, $t = 0,1,2,\dots$. Let $h_0>0$ and $0<a<1$. Define the sequence  $h_{t+1} = a h_t+\nu_t$. Show that $h_n$ is linearly converging.
----My Attempt:
Let $c:=\max\{a+\epsilon,b\}$ for any $\epsilon>0$. We  prove by induction that $h_t\leq \frac{c^t}{\epsilon}(h_0+\nu_0)$ for any $\epsilon>0$. It follows by the induction hypothesis that
\begin{equation}
\begin{aligned}
h_{t+1}&= a h_t +\nu_t \leq \frac{a}{\epsilon} c^t (h_0+\nu_0)+\frac{\epsilon}{\epsilon} c^t \nu_0\\
&\leq \frac{a}{\epsilon} c^t (h_0+\nu_0)+\frac{c-a}{\epsilon} c^t (h_0+\nu_0) = c^{t+1}(h_0+\nu_0)/\epsilon,
\end{aligned}
\end{equation}
where we used the fact that $c-a\geq \epsilon$ and $\nu_0 \leq h_0+\nu_0$.
----What I need:
I'd like an approach that does not introduce $\epsilon$. Any help is much appreciated.
 A: If you calculate $h_n$ for $n=0,1,2,3,4$, it’s easy to conjecture that
$$h_n=a^nh_0+\nu_0\sum_{k=0}^{n-1}a^kb^{n-1-k}$$
and prove it by induction. If $a\ne b$ it can then be written more compactly as
$$h_n=a^nh_0+\frac{(a^n-b^n)\nu_0}{a-b}=(h_0+c)a^n-cb^n$$
where $c=\frac{\nu_0}{a-b}$. If $a=b$, $h_n=h_0a^n+n\nu_0b^n$.
A: I like to make things telescope.
If
$h_{t+1} = a h_t+v_t
$
then,
dividing by $a^{t+1}$,
$\dfrac{h_{t+1}}{a^{t+1}}
= \dfrac{a h_t}{a^{t+1}}+\dfrac{v_t}{a^{t+1}}
= \dfrac{ h_t}{a^{t}}+\dfrac{v_t}{a^{t+1}}
$.
Note that
I am not worrying yet
about $v_t$.
Letting
$\dfrac{ h_t}{a^{t}}
=g_t
$,
this becomes
$g_{t+1}
= g_t+\dfrac{v_t}{a^{t+1}}
$.
Rearranging,
$g_{t+1}- g_t
=\dfrac{v_t}{a^{t+1}}
$.
This now telescopes
and we get
$\sum_{t=0}^{n-1}(g_{t+1}- g_t)
=\sum_{t=0}^{n-1}\dfrac{v_t}{a^{t+1}}
$
or
$g(n)-g(0)
=\sum_{t=0}^{n-1}\dfrac{v_t}{a^{t+1}}
$
so
$g(n)
=g(0)+\sum_{t=0}^{n-1}\dfrac{v_t}{a^{t+1}}
$
or
$\dfrac{ h_n}{a^{n}}
=g(0)+\sum_{t=0}^{n-1}\dfrac{v_t}{a^{t+1}}
=h(0)+\sum_{t=0}^{n-1}\dfrac{v_t}{a^{t+1}}
$
so
$h(n)
=a^nh(0)+a^n\sum_{t=0}^{n-1}\dfrac{v_t}{a^{t+1}}
=a^nh(0)+\sum_{t=0}^{n-1}a^{n-t-1}v_t
$.
Now we can put in the expression
for $v_t$
and get
$\begin{array}\\
h(n)
&=a^nh(0)+\sum_{t=0}^{n-1}a^{n-t-1}b^tv_0\\
&=a^nh(0)+a^{n-1}v_0\sum_{t=0}^{n-1}a^{-t}b^t\\
&=a^nh(0)+a^{n-1}v_0\sum_{t=0}^{n-1}(b/a)^t\\
&=a^nh(0)+a^{n-1}v_0\dfrac{1-(b/a)^n}{1-b/a}
\qquad\text{(standard geometric series)}\\
&=a^nh(0)+a^{n}v_0\dfrac{1-(b/a)^n}{a-b}\\
&=a^n\left(h(0)+v_0\dfrac{1-(b/a)^n}{a-b}\right)\\
\end{array}
$
