Approximating e with a guaranteed error

1. For some $\theta\in(0,1)$,
$$e=\sum_{k=0}^n\frac1{k!}+\frac{e^\theta}{(n+1)!}.$$

2. Error bound. Since $0<\theta<1$, $e^\theta<e<3$, so the error satisfies
$$0<e-\sum_{k=0}^n\frac1{k!}<\frac3{(n+1)!}.$$
- For $n=8$: $\frac3{9!}\approx8.3\cdot10^{-6}$, not small enough.
- For $n=9$: $\frac3{10!}\approx8.3\cdot10^{-7}<10^{-6}$.

So $n=9$ guarantees the precision (and it is the first $n$ for which this *bound* does).

3. $\sum_{k=0}^9\frac1{k!}=2.7182815255\dots$, while $e=2.7182818284\dots$; the actual error is $3.0\cdot10^{-7}$, indeed below the bound.

Remark. The factorials in the denominator make this series converge extremely fast: ten terms give six correct digits. The same reasoning is how calculators compute $e^x$.