The Lie product formula

Idea. Both $X_k$ and $Y_k$ equal $I+\frac{A+B}k$ up to $O(1/k^2)$; raising to the $k$-th power, $k$ errors of size $1/k^2$ add up to $O(1/k)\to0$.

Step 1: $\|X_k-Y_k\|\le C/k^2$. Write $e^{M}=I+M+r(M)$ with $r(M)=\sum_{m\ge2}M^m/m!$ and $\|r(M)\|\le\frac{\|M\|^2}2e^{\|M\|}$ (since $\sum_{m\ge2}x^m/m!\le\frac{x^2}2e^x$). Fix $a=\|A\|$, $b=\|B\|$ and let $k\ge1$. Then
$$X_k=\Bigl(I+\tfrac Ak+r_A\Bigr)\Bigl(I+\tfrac Bk+r_B\Bigr)=I+\tfrac{A+B}k+\Bigl(\tfrac{AB}{k^2}+r_A+r_B+\tfrac Akr_B+r_A\tfrac Bk+r_Ar_B\Bigr),$$
$$Y_k=I+\tfrac{A+B}k+r\bigl(\tfrac{A+B}k\bigr),$$
with $\|r_A\|\le\frac{a^2}{2k^2}e^a$, $\|r_B\|\le\frac{b^2}{2k^2}e^b$, $\|r(\tfrac{A+B}k)\|\le\frac{(a+b)^2}{2k^2}e^{a+b}$. All the correction terms are bounded by $C/k^2$ with $C$ depending only on $a,b$. Hence $\|X_k-Y_k\|\le C/k^2$.

Step 2: norms of $X_k,Y_k$. $\|X_k\|\le e^{a/k}e^{b/k}=e^{(a+b)/k}$ and $\|Y_k\|\le e^{\|A+B\|/k}\le e^{(a+b)/k}$. So both are bounded by $M_k=e^{(a+b)/k}$ (for the operator norm, or up to the harmless constant $\|I\|$).

Step 3: telescoping. The identity $X^k-Y^k=\sum_{j=0}^{k-1}X^{j}(X-Y)Y^{k-1-j}$ (check by expanding: consecutive terms cancel) gives
$$\|X_k^k-Y_k^k\|\le\sum_{j=0}^{k-1}\|X_k\|^j\|X_k-Y_k\|\|Y_k\|^{k-1-j}\le k\,M_k^{k-1}\cdot\frac C{k^2}\le\frac Ck\,e^{a+b}.$$
(using $M_k^{k-1}\le M_k^k=e^{a+b}$).

Conclusion. $\|(e^{A/k}e^{B/k})^k-e^{A+B}\|\le\frac{Ce^{a+b}}k\to0$.

Remark. The formula shows that although $e^Ae^B\ne e^{A+B}$ in general, the products of "infinitesimal steps" recover $e^{A+B}$. It is the mathematical core of Trotter-Kato splitting methods in numerical analysis and of the fact that the Lie algebra of a closed subgroup is a vector space (next problem).