Uniform convergence of matrix integral sequence I was given recursively defined: 
$$
M_k(t)=I+\int_{t_0}^tA(s)M_{k-1}(s)~ds
$$ 
and $M_0=I$ and that $A(t)$ is a matrix with entries that are continuous functions on $t_0\leq t\leq t_1$.
By induction we can get (I think) $$M_n=\sum\limits_{r=0}^n\left(\int_{t_0}^tA(s)~ds\right)^{\,r}.$$
Now, show that $M_n$ converges uniformly (componentwise) to some M (on the given interval). I really have no clue how to do this, any help is appreciated!
To me this seems to boil down to showing that $\int_{t_0}^tA(s)~ds$ is sufficiently small, $<1$ or something to use cauchy, but again, I don't see what given properties I could use!
 A: Unfortunately this formula 
$$
M_n(t)=I+\sum_{j=0}^n\left(\int_{t_0}^t A(s)\,ds\right)^{\!j}
$$
is not in general true. 
It is true if $A(s)A(t)=A(t)A(s)$, for all $s,t$.
In order to show that $M_n$ converges uniformly to some $M$, you basically need to follow the steps of the proof of Picard-Lindelöf.
So, subtracting
$$
M_n(t)=I+\int_{t_0}^t A(s)M_{n-1}(s)\,ds$$and$$
M_{n+1}(t)=I+\int_{t_0}^t A(s)M_n(s)\,ds\tag{1}
$$
we obtain
$$
\| M_{n+1}(t)-M_n(t)\| \le \int_{t_0}^t \|A(s)\|\|M_{n}(s)-M_{n-1}(s)\|\,ds\le 
a\int_{t_0}^t \|M_{n}(s)-M_{n-1}(s)\|\,ds,
$$
where $a=\max_{t\in I} \|A(t)\|$, and using the fact that 
$$
\|M_1(t)-M_0(t)\|\le \int_{t_0}^t \|A(s)\|\,ds\le a(t-t_0),
$$
you inductively obtain that
$$
\| M_{n+1}(t)-M_n(t)\|\le \frac { a^{n+1}(t-t_0)^{n+1}} {(n+1)!}\le 
\frac { a^{n+1}(t_1-t_0)^{n+1}} {(n+1)!},
$$
which by virtue of Weierstrass criterion implies that $M_n(t)$ converges usiformly to say $M(t)$, and its limit satisfies the IVP
$$
X'=A(t)X, \quad X(t_0)=I.
$$
Due to the following fact, letting $n\to\infty$ in $(1)$ we obtain
$$
M(t)=I+\int_0^t A(s)M(s)\,ds.
$$
Then $M(0)=I$, and $M'(t)=A(t)M(t)$.
A: This is the same as Yiorgos' answer, but I was writing mine when I discovered his...
Choose any submultiplicative norm on the space of matrices (i.e. $\Vert AB\Vert\leq\Vert A\Vert\,\Vert B\Vert$). Also, for any matrix-valued function $\Phi$ on $[t_0,t_1]$, set $\Vert\Phi\Vert_\infty:=\sup \{ \Vert \Phi(t)\Vert;\, t\in [t_0,t_1]\}$.
Since $A$ is continuous, it is bounded on $[t_0,t_1]$, say $\Vert A\Vert_\infty=M<\infty$.
Let us show by induction that 
$$\Vert M_{n+1}-M_n(t)\Vert \leq \frac{M^{n+1}}{(n+1)!}  (t-t_0)^{n+1}$$
for every $n\geq 0$ and all $t\in [t_0,t_1]$.
For $n=0$ we have 
$$\Vert M_1(t)-M_0(t)\Vert =\left\Vert \int_{t_0}^t A(s)\, ds\right\Vert\leq M\, (t-t_0)\, . $$
Assume the inequality has been proved for some $n$. Then 
\begin{eqnarray}
\Vert M_{n+2}(t)-M_{n+1}(t)\Vert&=&\left\Vert \int_{t_0}^t A(s)(M_{n+1}(s)-M_n(s))\, ds \right\Vert\\
&\leq&\int_{t_0}^t M\times \frac{M^{n+1}}{(n+1)!} (s-t_0)^{n+1}\, ds\\
&=&\frac{M^{n+2}}{(n+2)!}\, (t-t_0)^{n+2}\cdot
\end{eqnarray}
It follows that $\Vert M_{n+1}-M_n\Vert_\infty\leq \frac{M^{n+1}}{(n+1)!}  (t_1-t_0)^{n+1}$ for all $n$. Hence, the series $\sum\Vert M_{n+1}-M_n\Vert_\infty$ is convergent, and so the sequence $(M_n)$ is uniformly convergent on $[t_0,t_1]$.
A: Just to add my two cents. The matrix solution $M(t)$ of the pertinent integral equation plays a vital role in defining the solution of Linear Time Varying Systems (LTVs). In fact, the solution map of an LTV can be defined as $M(t)M(t_0)^{-1}$, where $t_0$ is the initial time.
This obviously alludes to the important fact that $M(t)$ is non-singular.This can be easily shown using the small gain theorem. 
