why can i differentiate this term-by-term?

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What's the best way to justify the following computation:

For $A, B$ symmetric real matrices,

$$\frac{d}{dt}|_{t=0}e^{A+tB}= \frac{d}{dt}|_{t=0}(1+(A+tB)+\frac{1}{2!}(A+tB)^2+...) = (B+\frac{1}{2!}(BA+AB)+...+\frac{1}{(n+1)!}\sum _{k=0}^n A^kBA^{n-k}+...)$$

The only thing I don't understand is the how to justify term-by-term differentiation. In the case of an expression like $\frac{d}{dt}e^{tB}$ it is easy that each coordinate is a power series in t with infinite radius of convergence. Is that what I should do here or is there an easier way?

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The series $\sum_{n=0}^\infty \frac 1{n!}(A+tB)^n$ and its termwise derivative $\sum_{n=0}^\infty \frac 1{n!} \frac{d}{dt} (A+tB)^n$ converge absolutely and uniformly on compact sets. Hence the limits (differentiation and series summing) can be exchanged, giving $$ \frac{d}{dt}\sum_{n=0}^\infty \frac 1{n!} (A+tB)^n = \sum_{n=0}^\infty \frac 1{n!} \frac{d}{dt} (A+tB)^n $$