TITLE: At what rate do these common definitions converge to the constant $e=2.718...$?
QUESTION [7 upvotes]: $e=2.718\ldots$ is often defined to be $$\lim_{n\to\infty} (1+1/n)^n$$ (from continuously compounded interest) and $$\lim_{n\to\infty}\sum_{k=0}^n \frac{1}{k!}$$ (from Taylor series). My question is, what are the respective rates of convergence for each of these? For example, is the error from $e$ in $O(\frac{1}{n})$? Or is the number of digits of accuracy in $O(n)$? If it's not too involved, a basic overview of a proof would also be appreciated.
And just for bonus points, since I'm curious, what is the state-of-the-art algorithm for calculating digits of $e$?

REPLY [5 votes]: In the first case,
\begin{align*}
\left(1 + \frac{1}{n}\right)^n
   &= \exp \left[n \log \left(1 + \frac{1}{n}\right)\right]
    = \exp \left[1 - \frac{1}{2n} + O\left(n^{-2}\right)\right]
    = e - \frac{e}{2n} + O(n^{-2})
\end{align*}
as $n\to\infty$. In the second case,
\begin{align*}
\sum_{k \leq n} \frac{1}{k!} - e
   = \sum_{k > n} \frac{1}{k!} 
   &= \sum_{k > 0} \frac{1}{(k + n)!}
   \leq \sum_{k > 0} \frac{1}{n^k\, n!}
   = \frac{1}{n!(n-1)}
   = O\left(\frac{1}{n!}\right)
\end{align*}