[proofplan]
We write the logarithm series explicitly and prove absolute convergence by comparison with a scalar series. Submultiplicativity gives the power estimate $\|A^m\|_*\le \|A\|_*^m$ for every $m\ge 1$. Since $r:=\|A\|_*$ satisfies $0\le r<1$, the scalar majorant $\sum_{m=1}^{\infty}r^m/m$ converges. The partial sums are therefore Cauchy in the finite-dimensional complete normed space $M(n,\mathbb C)$, so the logarithm series converges absolutely.
[/proofplan]
[step:Estimate every matrix power by submultiplicativity]
Let $r:=\|A\|_*$. Since $\|\cdot\|_*$ is a norm, $r\ge 0$, and by hypothesis $r<1$.
We prove by induction that
\begin{align*}
\|A^m\|_* \le r^m
\end{align*}
for every $m\in\mathbb N$. For $m=1$, this is the identity $\|A\|_*=r$. Suppose the estimate holds for some $m\in\mathbb N$. By submultiplicativity applied to $A^m$ and $A$,
\begin{align*}
\|A^{m+1}\|_*=\|A^mA\|_* \le \|A^m\|_*\|A\|_* \le r^m r = r^{m+1}.
\end{align*}
The induction is complete.
[/step]
[step:Compare the absolute-value series with a convergent scalar series]
For each $m\in\mathbb N$, homogeneity of the norm and the preceding power estimate give
\begin{align*}
\left\|\frac{(-1)^{m+1}A^m}{m}\right\|_*=\frac{\|A^m\|_*}{m}\le \frac{r^m}{m}.
\end{align*}
Since $m\ge 1$, we have $1/m\le 1$, and hence
\begin{align*}
0\le \frac{r^m}{m}\le r^m.
\end{align*}
The geometric series $\sum_{m=1}^{\infty}r^m$ converges because $0\le r<1$. Therefore, by comparison,
\begin{align*}
\sum_{m=1}^{\infty}\frac{r^m}{m}<\infty.
\end{align*}
Consequently,
\begin{align*}
\sum_{m=1}^{\infty}\left\|\frac{(-1)^{m+1}A^m}{m}\right\|_*<\infty.
\end{align*}
[guided]
The goal is to prove absolute convergence, so we must control the scalar series formed by the norms of the terms:
\begin{align*}
\sum_{m=1}^{\infty}\left\|\frac{(-1)^{m+1}A^m}{m}\right\|_*.
\end{align*}
The sign $(-1)^{m+1}$ has no effect on the norm, because $|(-1)^{m+1}|=1$. Homogeneity of the norm gives, for every $m\in\mathbb N$,
\begin{align*}
\left\|\frac{(-1)^{m+1}A^m}{m}\right\|_*=\frac{\|A^m\|_*}{m}.
\end{align*}
The useful comparison comes from submultiplicativity. With $r:=\|A\|_*$, the previous step proved
\begin{align*}
\|A^m\|_* \le r^m
\end{align*}
for every $m\in\mathbb N$. Therefore
\begin{align*}
\left\|\frac{(-1)^{m+1}A^m}{m}\right\|_* \le \frac{r^m}{m}.
\end{align*}
Now we need to know that the scalar majorant converges. Since $m\ge 1$, we have $1/m\le 1$, so
\begin{align*}
0\le \frac{r^m}{m}\le r^m.
\end{align*}
The geometric series $\sum_{m=1}^{\infty}r^m$ converges because $0\le r<1$. By the [comparison test](/theorems/173) for non-negative real series,
\begin{align*}
\sum_{m=1}^{\infty}\frac{r^m}{m}<\infty.
\end{align*}
Applying the same comparison to the norm series gives
\begin{align*}
\sum_{m=1}^{\infty}\left\|\frac{(-1)^{m+1}A^m}{m}\right\|_*<\infty.
\end{align*}
This is exactly absolute convergence of the matrix logarithm series with respect to the norm $\|\cdot\|_*$.
[/guided]
[/step]
[step:Use completeness of the matrix space to obtain convergence]
For $N\in\mathbb N$, define the $N$th partial sum $S_N\in M(n,\mathbb C)$ by
\begin{align*}
S_N:=\sum_{m=1}^{N}\frac{(-1)^{m+1}A^m}{m}.
\end{align*}
If $N>K$, then the triangle inequality and the estimate from the preceding step give
\begin{align*}
\|S_N-S_K\|_* \le \sum_{m=K+1}^{N}\left\|\frac{(-1)^{m+1}A^m}{m}\right\|_*.
\end{align*}
The right-hand side tends to $0$ as $K,N\to\infty$ because the norm series converges. Thus $(S_N)_{N\in\mathbb N}$ is Cauchy in $(M(n,\mathbb C),\|\cdot\|_*)$.
The [vector space](/page/Vector%20Space) $M(n,\mathbb C)$ is finite-dimensional over $\mathbb C$, so it is complete for every norm. Hence $(S_N)_{N\in\mathbb N}$ converges to an element of $M(n,\mathbb C)$. Since its norm series also converges, the logarithm [power series](/page/Power%20Series) converges absolutely. This proves the theorem.
[/step]
