Why can we distribute the transpose operation on an **infinite** sum of matrices? The matrix exponential is defined by the sum of an infinite series. To compute the transpose of this exponential:
$(e^M)^T = (\sum_{i=0}^\infty \frac{M^i}{i!})^T = \sum_{i=0}^\infty \frac{(M^i)^T}{i!}$
In general, I do not like to apply a rule an infinite number of times. Sometimes it may lead to wrong conclusions. So why the distributivity of the transpose operation can be applied in this case? Is it because the fact that the sum converges? So, I'm looking for any reference to a formal proof for that property.
Once this is accepted, it is obvious that $(M^i)^T = (M^T)^i$. The variable $i$ can be very large but it is finite. By induction, this is true for all $i$. Thus, $(e^{M})^T = e^{M^T}$.
Thank you for your help.
 A: All you need to check is that the transpose commutes with taking a limit. This is really easy to see: What is the limit of a sequence of matrices? It is just the limit of every entry. The transpose just permutes the entries, it does nothing to the individual limits. Let me illustrate this in the $2 \times 2$ case. Let $(A^{(n)})_{n \in \mathbb{N}}$ be a sequence of $2 \times 2$ matrices. Then
$$\left(\lim\limits_{n \to \infty} A^{(n)}\right)^T = \left(\lim\limits_{n \to \infty} \begin{pmatrix} A^{(n)}_{11} & A^{(n)}_{12} \\ A^{(n)}_{21} & A^{(n)}_{22} \end{pmatrix}\right)^T = \begin{pmatrix} \lim\limits_{n \to \infty} A^{(n)}_{11} & \lim\limits_{n \to \infty} A^{(n)}_{12} \\ \lim\limits_{n \to \infty} A^{(n)}_{21} & \lim\limits_{n \to \infty} A^{(n)}_{22} \end{pmatrix}^T = \begin{pmatrix} \lim\limits_{n \to \infty} A^{(n)}_{11} & \lim\limits_{n \to \infty} A^{(n)}_{21} \\ \lim\limits_{n \to \infty} A^{(n)}_{12} & \lim\limits_{n \to \infty} A^{(n)}_{22} \end{pmatrix} = \lim\limits_{n \to \infty} \begin{pmatrix} A^{(n)}_{11} & A^{(n)}_{21} \\ A^{(n)}_{12} & A^{(n)}_{22} \end{pmatrix} = \lim\limits_{n \to \infty} (A^{(n)})^T.$$
