Operator Exponential $e^A e^B = e^{A+B}$

In any Banach algebra, the Cauchy product of two absolutely convergent series is absolutely convergent, and with the expected sum. That is, if $\sum_{j=0}^\infty \|a_j\|<\infty$ and $\sum_{k=0}^\infty \|b_k\|<\infty$, then defining $c_m = \sum_{j+k=m}a_jb_k$, we get an absolutely convergent series and $$ \sum_{m=0}^\infty c_m = \left(\sum_{j=0}^\infty a_j \right)\left(\sum_{k=0}^\infty b_k\right) \tag1 $$ The proof is literally the same as the proof for real/complex numbers. One doesn't even need $a_j$ and $b_k$ to commute to have (1), because they are always multiplied in the same order. But here is a proof anyway.

Proof : Let $A_n$, $B_n$, $C_n$ denote partial sums over indices $0,\dots,n$. Consider the difference $A_nB_n-C_n$. It consists of all terms $a_j b_k$ where $ j,k\le n$ and $j+k>n$. By the triangle inequality, it suffices to prove that $$ \sum_{j,k\le n, \ j+k>n} \|a_j\| \,\|b_k\| \tag2 $$ is small when $n$ is large. Since either $j$ or $k$ has to be $>n/2$, we can estimate (2) from above by $$ \sum_{j\le n,\ n/2<k\le n} \|a_j\| \,\|b_k\| + \sum_{n/2<j\le n,\ k\le n} \|a_j\| \,\|b_k\| \tag 3$$ which can be rewritten as $$ \sum_{j\le n} \|a_j\| \sum_{n/2<k\le n} \|b_k\| + \sum_{k\le n}\|b_k\| \sum_{n/2<j\le n} \|a_j\| \tag 4$$ As $n\to \infty$, the first factor in each product stays bounded while the second factor goes to zero.