## On the convergence of formal power series of matrices

Let $G(x) = \sum_{k \in \mathbb{N}} \alpha_kx^k$ be a formal power series over $\mathbb{C}$ with radius of convergence $R$. Let $||A|| = \sum_{i,j} |a_{i,j}|$ be the matrix norm introduced in this previous exercise.

Given a matrix $A$ over $\mathbb{C}$, we can construct a sequence of matrices by taking the $N$th partial sum of $G(A)$. That is, $G_N(A) = \sum_{k = 0}^N a_kA^k$. This gives us $n^2$ sequences $\{c_{i,j}^N\}$ where $c_{i,j}^N$ is the $(i,j)$ entry of $G_N(A)$. Suppose $c_{i,j}^N$ converges to $c_{i,j}$ for each $(i,j)$, and let $C = [c_{i,j}]$. In this situation, we say that $G_N(A)$ converges to $C$, and that $G(A) = C$. (In other words, $G_N(A)$ converges precisely when each entrywise sequence $G_N(A)_{i,j}$ converges.)

1. Prove that if $||A|| \leq R$, then $G_N(A)$ converges.
2. Deduce that for all matrices $A$, the following power series converge: $\mathsf{sin}(A) = \sum_{k \in \mathbb{N}} \dfrac{(-1)^k}{(2k+1)!}A^{2k+1}$, $\mathsf{cos}(A) = \sum_{k \in \mathbb{N}} \dfrac{(-1)^k}{(2k)!}A^{2k}$, and $\mathsf{exp}(A) = \sum_{k \in \mathbb{N}} \dfrac{1}{k!} A^k$.

[Disclaimer: My facility with even simple analytical concepts is laughably bad, but I’m going to give this a shot. Please let me know what’s wrong with this solution.]

We begin with a lemma.

Lemma: For all $(i,j)$, $(A^k)_{i,j} \leq ||A||^k$, where the subscripts denote taking the $(i,j)$ entry. Proof: By the definition of matrix multiplication, we have $A^k = [\sum_{t \in n^{k-1}} \prod_{i=0}^{k-1} a_{t(i), t(i+1)}]$, where $t(0) = i$ and $t(k+1) = j$. (I’m abusing the notation a bit here; $t$ is an element of $\{1,\ldots,n\}^{k-1}$, which we think of as a function on $\{1,\ldots,k-1\}$.) Now $|A^k_{i,j}| \leq \sum_{t \in n^{k-1}} \prod_{i=0}^{k-1} |a_{t(i), t(i+1)}|$ by the triangle inequality. Note that $||A||^k$ is the sum of all possible $k$-fold products of (absolute values of) entries of $A$, and that we have $|A^k_{i,j}|$ bounded above by a sum of some distinct $k$-fold products of (absolute values of) entries of $A$. In particular, $|A^k_{i,j}| \leq ||A||^k$ since the missing terms are all positive. Thus $|\alpha_k A^k_{i,j}| = |(\alpha_k A^k)_{i,j}| \leq |\alpha_k| ||A||^k$.

Let us define the formal power series $|G|(x)$ by $|G|(x) = \sum |\alpha_k| x^k$. What we have shown (I think) is that $\sum_{k=0}^N |\alpha_k A^k_{i,j}| \leq |G|_N(||A||)$, using the triangle inequality.

Now recall, by the Cauchy-Hadamard theorem, that the radius of convergence of $G(x)$ satisfies $1/R = \mathsf{lim\ sup}_{k \rightarrow \infty} |\alpha_k|^{1/k}$. In particular, $|G|$ has the same radius of convergence as $G$. So in fact the sequence $|G|_N(||A||)$ converges. Now the sequence $\sum_{k=0}^N |\alpha_k A^k_{i,j}|$ is bounded and monotone increasing, and so must also converge. That is, $\sum_{k=0}^N (\alpha_k A^k)_{i,j} = G_N(A)_{i,j}$ is absolutely convergent, and so is convergent. Thus $G(A)$ converges.

Now we will address the convergence of the series $\mathsf{sin}$. Recall that the radius of convergence $R$ of $\mathsf{sin}$ satisfies $R^{-1} = \mathsf{lim\ sup}_{k \rightarrow \infty} |\alpha_k|^{1/k}$ $\mathsf{lim}_{k \rightarrow \infty} \mathsf{sup}_{n \geq k} |\alpha_k|^{1/k}$. Now $\alpha_k = 0$ if $k$ is even and $(-1)^t/k!$ if $k = 2t+1$ is odd. So $|\alpha_k|^{1/k} = 0$ if $k$ is even and $|1/k!|^{1/k} = 1/(k!)^{1/k} > 0$ if $k$ is odd.

Brief aside: Suppose $k \geq 1$. Certainly $1 \leq k! < (k+1)^k$, so that $(k!)^{1/k} < k+1$. Now $1 \leq (k!)^{1 + 1/k} < (k+1)!$, so that $(k!)^{1/k} < ((k+1)!)^{1/(k+1)}$. In particular, if $n > k$, then $(k!)^{1/k} < (n!)^{1/n}$.

By our brief aside, we have that $\mathsf{sup}_{n \geq k} |\alpha_k|^{1/k} = 1/(k!)^{1/k}$ if $k$ is odd, and $1/((k+1)!)^{1/(k+1)}$ if $k$ is even. So (skipping redundant terms) we have $\mathsf{sin}(x)$ satisfies $R^{-1} = \mathsf{lim}_{k \rightarrow \infty} 1/(k!)^{1/k}$. We know from the Brief Aside that this sequence is monotone decreasing. Now let $\varepsilon > 0$. Consider now the sequence $\varepsilon$, $2 \varepsilon$, $3 \varepsilon$, et cetera. only finitely many terms of this sequence are less than or equal to 1. So there must exist some $k$ such that the product $\prod_{i=1}^k \varepsilon i$ is greater than 1. Now $\varepsilon^k k! > 1$, so that $\varepsilon > 1/(k!)^{1/k}$. Thus the sequence $1/(k!)^{1/k}$ converges to 0, and so the radius of convergence of $\mathsf{sin}(x)$ is $\infty$.

By a similar argument, the radii of convergence for $\mathsf{cos}(X)$ and $\mathsf{exp}(X)$ are also $\infty$.