Evaluating the precision in the calculation of $\mathrm{e}$ I'm calculating $\mathrm{e}$ using a computer like this:
$$
\mathrm{e} \approx \sum\limits_{i=0}^n {1\over i!}
$$
I'm storing it as a rational number. 
I was wondering, if I write down my rational number as a decimal number, could I determine, 
how many digits after the decimal point are correct for a given value of $n$?
 A: The tail of the series is bounded above by a geometric series:
\begin{align}
\sum_{i=n+1}^\infty \frac{1}{i!} \le \frac{1}{(n+1)!}\sum_{i=0}^\infty \frac{1}{(n+1)^i}.
\end{align}
It's easy to find the sum of that series, so you get an upper bound on $e$.
The lower bound comes from stopping after finitely many terms.
A: Since the factorials grow so fast, the first term you ignore is a very good estimate for the error.  So if you sum up through $n=10$, the first ignored term is $\frac 1{11!}\approx 2.5\cdot 10^{-8}$  The next is a factor $12$ smaller, so using the first as your error estimate is pretty good.
A: Using the Taylor remainder formula you get that, for some $\xi\in (0,1)$
$$
0<\mathrm{e}-\sum_{i=0}^n\frac{1}{n!}=\frac{\mathrm{e}^\xi}{(n+1)!}<\frac{3}{(n+1)!}.
$$
Thus the error is less than $\frac{3}{(n+1)!}$.
A: As yet another possibility, if you calculate
$$\frac1e = \sum_{i=0}^\infty \frac{(-1)^i}{i!} $$
then you have an alternating series, so the true value of $1/e$ is strictly between any two successive partial sums, which you can then invert and represent in decimal. Any digits they agree on are certain.
