Limiting value of iteration $x(k+1) = A x(k) + B u(k)$ for summable $u(k)$ A matrix $A$ is known to converge such that $\lim_{k\rightarrow \infty} A^k  = \bar{A} \neq 0$. We have an iteration defined as
$$x(k+1) = A x(k) + B u(k), \ \ k\in \mathbb{Z}_+.$$
$\{u(k), k=0,1,..\}$ is an infinite sequence. Under what kind of sequence $\{u(k)\}$, will $x(k)$ converge as $k \rightarrow \infty$?
I suspect the sequence converges if $\{u(k)\}$ is absolutely summable and, in that case, converges to $\bar{A}x_0 + \bar{A} B \sum_{k=0}^\infty u(k)$ where $x_0 = x(0)$. However, I am not sure how to prove it. Would really appreciate any help on this.
 A: A sufficient condition is summability of $u$.
The solution to the difference equation is $x_k = A^k x_0 + \sum_{i=0}^k A^{k-i} B u_i$.
Since $A^k \to \overline{A}$, we see that $\|A^k\| \le M$ for some $M$ and for all $k \ge 0$. (It follows that $\|\overline{A}\| \le M$.)
We have$\sum_{i=0}^k \|A^{k-i}\| \|B\| \|u_i\| \le M \|B\| \sum_{i=0}^k \|u_i\| \le M \|B\| \sum_{i=0}^\infty \|u_i\|$, hence $x_1 = \lim_{k \to \infty} \sum_{i=0}^k A^{k-i} B u_i$
exists.
It follows that $x_k \to \overline{A} x_0 + x_1$.
If $u$ is summable, then $\|u_k \| \le M'$ for some $M'$ and for all $k \ge 0$.
Let $ \epsilon >0$ and choose $N$ such that  $2 M \|B\| \sum_{i>N} \|u_i\| < \frac{1}{2} \epsilon$. Now choose $N' \ge N$ such that for $k \ge N'$, $\|B\| M' \max(\|A^k-\overline{A}\|, \|A^{k-1}-\overline{A}\|,..., \|A^{k-N}-\overline{A}\|) < \frac{1}{2(N+1)} \epsilon$.
If $k \ge N'$, then
\begin{eqnarray}
\| \sum_{i=0}^k (A^{k-i} - \overline{A}) B u_i \| &\le& \| \sum_{i=0}^N (A^{k-i} - \overline{A}) B u_i \| + \| \sum_{i>N} (A^{k-i} - \overline{A}) B u_i \| \\
&\le& \|B\|M' \sum_{i=0}^N \| A^{k-i} - \overline{A} \| + 2 M \|B\| \sum_{i>N} \|u_i\| \\
&<& (N+1)\frac{1}{2(N+1)} \epsilon + \frac{1}{2} \epsilon \\
&=& \epsilon
\end{eqnarray}
It follows that $x_1 = \lim_{k \to \infty} \sum_{i=0}^k A^{k-i} B u_i
= \lim_{k \to \infty} \sum_{i=0}^k \overline{A} B u_i  = \overline{A} B \sum_{i=0}^\infty  u_i$, and so
$\lim_{k \to \infty} x_k = \overline{A} x_0 + \overline{A} B \sum_{i=0}^\infty  u_i$.
