Compute e with accuracy 10^-9

Compute $e$ with accuracy $10^{-9}$.

Step 1: Series representation

Setting $x=1$ in the Maclaurin series for $e^x$: \[e = \sum_{k=0}^n\frac1{k!}+R_n, \qquad R_n=\frac{e^\xi}{(n+1)!} \lt \frac3{(n+1)!},\ \ 0\lt \xi\lt 1. \tag{1}\]

Step 2: Choose $n$ so that $R_n\lt 10^{-9}$

Testing successive factorials: $12!\approx4.79\times10^8\lt 3\times10^9$, but $13!\approx6.23\times10^9>3\times10^9$. So $n=12$ suffices: \[R_{12} \lt \frac3{13!} \approx 4.8\times10^{-10} \lt 10^{-9}. \tag{2}\]

Step 3: Sum the thirteen terms

\[ S_{12} = \sum_{k=0}^{12}\frac1{k!} \approx 2.718281828286. \tag{3} \]

Step 4: Verification

\[S_{12}=2.71828182828617\ldots,\ e=2.71828182845905\ldots,\ |S_{12}-e|\approx1.73\times10^{-10}\lt 10^{-9}\ \checkmark\]