Let $A$ be a bounded linear operator on a normed vector space $X$.
Then, how do I show rigorously that:
$$\lim_{h\to0} \sum_{n=0}^\infty\dfrac{1}{(n+1)!}A^nh^n \to I $$ in the operator norm?
What I tried was to naively write out the sum as
$$I + \dfrac{1}{2!}Ah + \dfrac{1}{3!}A^2h^2 \dots $$
and since $h\to 0$ I know that each term that has $h$ in it vanishes, but how do I show rigoursly that the infinite sum of those vanishing terms vanish to zero?
Hint. Notice that $$ \left\|I-\sum_{n=0}^\infty \frac{1}{(n+1)!}A^nh^n\right\| \le \sum_{n=1}^\infty \left\|\frac{1}{(n+1)!}A^nh^n\right\| = \sum_{n=1}^\infty \frac{h^n}{(n+1)!}\left\|A^n\right\| \le \sum_{n=1}^\infty \frac{h^n}{(n+1)!}\left\|A\right\|^n $$ Thus if $\sum_{n=1}^\infty \frac{h^n}{(n+1)!}\left\|A\right\|^n \to 0$ as $h\to 0$ then $\sum_{n=0}^\infty \frac{1}{(n+1)!}A^nh^n \to I$ as $h\to0$. You should be able to prove $\sum_{n=1}^\infty \frac{h^n}{(n+1)!}\left\|A\right\|^n \to 0$ as $h\to 0$ relatively easily.