Exponentials of operators If $\{T_n\}$ converges in operator norm to $T$ does it follow that $\exp(T_n)$ converges to $\exp(T)$ at least in finite dimensions? 
I can handle this when the operators commute but not in general.
 A: $$
    \exp(T_n)-\exp(T)=\exp((1-t)T_n)\exp(tT)|_{t=0}^{1} \\
    = \int_{0}^{1}-\exp((1-t)T_n)T_n\exp(tT)+\exp((1-t)T_n)T\exp(tT)dt \\
    = \int_{0}^{1}\exp((1-t)T_n)(T-T_n)\exp(tT)dt
$$
Therefore,
$$
      \|\exp(T_n)-\exp(T)\| \le \int_{0}^{1}\exp((1-t)\|T_n\|)\|T-T_n\|\exp(t\|T\|)dt.
$$
A: The estimates I had for $||A^n-B^n||$ don't hold for non-commuting operators but a simple application of Dominated Convergence Theorem proves continuity of the exponential map in the operator norm, even in infinite dimensions. Anyway, thanks for the comments. 
A: Let $S = T_n-T$ so that $\|S_n\| \to 0$ 
$$\exp(T+S_n)-\exp(T) = \sum_{k=1}^\infty \frac{1}{k!}((T+S_n)^k-T^k)$$
Use the non-commutative binomial sum
$$(T+S_n)^k = \sum_{a \in \{0,1\}^k } \prod_{i=1}^k T^{a_i} S_n^{1-a_i}$$
For $\|S_n\| \le \|T\|$
$$ \|(T+S_n)^k -T^k\| \le \sum_{a \in \{0,1\}^k,a \ne \{1\}^k } \|\prod_{i=1}^k T^{a_i} S_n^{1-a_i}\|\quad \le  \quad 2^k \|S_n\|\|T\|^{k-1}$$
$$\|\exp(T+S_n)-\exp(T)\| \le \|S_n\| \sum_{k=1}^\infty \frac{2^k \|T\|^{k-1}}{k!}=
\|S_n\| \frac{e^{2\|T\|}-1}{\|T\|}$$
