Question
Suppose $$f(x) = \sum_{k=0}^{\infty} c_k x^k, c_k \in \mathbb{R},$$ whenever $\lvert x \lvert < r$, where $r \in \mathbb{R}$ is the radius of convergence.
Similarly, the series expansion of a matrix function $f(A)$ is the corresponding series that is $$f(A) = \sum_{k=0}^{\infty} c_k A^k, c_k \in \mathbb{R},$$ where $A^0 = \mathbb{I}$.
Moreover, for a diagonalisable matrix $A$, we can use the series expansion of $f(A)$ (when it converges) to write $$f(A) = P \begin{bmatrix} f(\lambda_1) & \cdots & 0 \\ \vdots & \ddots & \vdots \\ 0 & \cdots & f(\lambda_n) \end{bmatrix} P^{-1},$$ where $\lambda_1, \cdots, \lambda_n$ are eigenvalues of $A$.
What conditions do the eigenvalues of $A$ need to satisfy for the power series expansion of the matrix function $f(A)$ to converge?
Hint : Think about what it means for f(x) to converge.
My answer
We know that, in order for $f(A)$ to converge, $$P \begin{bmatrix} f(\lambda_1) & \cdots & 0 \\ \vdots & \ddots & \vdots \\ 0 & \cdots & f(\lambda_n) \end{bmatrix} P^{-1}$$ must converge.
It is trivial to see that both $P$ and $P^{-1}$ are always convergent, so the problem reduces to checking the convergence of $$\begin{bmatrix} f(\lambda_1) & \cdots & 0 \\ \vdots & \ddots & \vdots \\ 0 & \cdots & f(\lambda_n) \end{bmatrix},$$ where $f(\lambda)$ takes the form of a geometric series $\sum_{k=0}^{\infty} a r^k$ which converges when $\lvert r \lvert < 1$. Thus, it follows that, for $f(\lambda)$ to converge, we need $\lvert \lambda \lvert < 1$ and so, for $f(A)$ to converge, we need $\lvert \lambda_i \lvert < 1\ \forall\ i \in \{1, \dots, n\}.$
Whilst discussing this with my professor, I think he asked me to consider the convergence for a general series and not just the geometric series, but then he proceeded to retract his words and say that my working was alright. Thus, I would like to check with the community here whether my answer is correct. If not, then any intuitive explanations or suggestions will be greatly appreciated!