Convergence of two nested geometric sequences

Question

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.

Answer 1

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 $a=b$, $h_n=h_0a^n+n\nu_0b^n$.

Answer 2

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} $