TITLE: Calculating $e$
QUESTION [7 upvotes]: If I calculate $e$ using the following formula.
$$e = \sum_{k=0}^{\infty}{\frac{1}{k!}}$$
Is it possible to predict how many correct decimal places I get when I stop summing at $n$ terms?

REPLY [5 votes]: In this answer, it is shown, by comparison to a geometric series, that
$$
0\le n!\left(e-\sum_{k=0}^n\frac1{k!}\right)\le\frac1n
$$
Therefore, the error after $n+1$ terms is at most $\frac1{nn!}$ .
To $n$ decimal places:
When asking for a number to $n$ decimal places, there are two common meanings 

the error is less than $\frac12\times10^{-n}$.
the value is correct when rounded to $n$ decimal places. As has been pointed out, if a number is very close to $10^{-n}\left(\mathbb{Z}+\frac12\right)$, rounding to $n$ decimal places might require computing more decimal places to know the actual $n^{\mathrm{th}}$ digit of the rounded number. This is not as easy to use as meaning 1, so it is not as commonly used.