[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*}[/step]

[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: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]