Let $f_n = \sum _{k=0} ^{n-1} \frac{(-1)^k(\ln a)^k}{k!}, 0<a<1$. My intuition says this converges very rapidly to $1/a$, so I need to compare it to a nice sequence. My candidate was $b_n = \left(\frac{1}{a}\right) \left(1-\frac{1}{n^2}\right)$.
MATLAB does indeed confirm that $f_n$ approaches $1/a$ much faster than $b_n$, i.e. for some $N$, $\forall n\geq N$ we have that $f_n>b_n$. I just can't seem to prove it on paper.
Any ideas or hints?
Here's a start:
$$\left | e^x - \sum_{k=0}^{N-1} \frac{x^k}{k!} \right | \leq \sum_{k=N}^\infty \left | \frac{x^k}{k!} \right | \leq \sum_{k=N}^\infty \left | \frac{x^k}{N!N^{k-N}} \right | = \frac{1}{N! N^{-N}} \sum_{k=N}^\infty \left | \frac{x^k}{N^k} \right |.$$
Now the last sum is well approximated by $\frac{x^N}{N^N}$, if $N$ is sufficiently large. (A useful heuristic: a convergent geometric series is well-approximated by its largest term.) So the overall quantity is well approximated by $\frac{x^N}{N!}$. Now fill in the details.