# Integer values of $(\frac{19}{7}-\sum_{k=0}^N \frac{1}{k!})^{-1}$

The approximation for $e$ involved in An integral for $2\pi+e-9$ exactly matches a truncation of factorial reciprocal series

$$\sum_{k=0}^5 \frac{1}{k!} =\frac{163}{60}$$

In contrast, the difference

$$\frac{19}{7}-\sum_{k=0}^N \frac{1}{k!}$$ never cancels but it is a unit fraction for $N=3,4,5,7$.

$$\frac{19}{7}-\sum_{k=0}^3 \frac{1}{k!}=\frac{1}{21}=\frac{2}{6·7}$$

$$\frac{19}{7}-\sum_{k=0}^4 \frac{1}{k!}=\frac{1}{168}=\frac{3}{7·8·9}$$

$$\frac{19}{7}-\sum_{k=0}^5 \frac{1}{k!}=-\frac{1}{420}=-\frac{4}{5·6·7·8}$$

$$\frac{19}{7}-\sum_{k=0}^7 \frac{1}{k!}=-\frac{1}{252}=-\frac{2}{7·8·9}$$

Is this list complete?

• At least, up to $N=1000$, this is true. – Claude Leibovici Jun 2 '17 at 8:03

For ease of notation I'll denoting $\sum_{k=0}^N \frac{1}{k!}$ as $e_N$ for now.
Note that $(19/7 - e)^{-1} \approx -250.24...$, or in particular it's somewhere in-between $-250$ and $-251$. So if we can show that for $N \ge 8$ the sum $(\frac{19}{7}-e_N)^{-1}$ is in between $-250$ and $-251$ then from that point it can't have any more integer values.
Since $-1/250 < 19/7 - e$ it is enough to show that for $N \ge 8$ we have that $19/7 - e_{N} < -1/251$, and since $e_{N}$ is montonically increasing we only need to show this for $N=8$.
Now we know that $\frac{19}{7} - e_{7} = -1/252$, and $$1/252 - 1/251 = 1/(251 \cdot 252) = 1 / 63252.$$ Since $8! = 40320 < (251 \cdot 252)$ we find that $$\frac{19}{7} - e_{8} = \frac{19}{7} - e_{7} - 1/8! = -1/252 - 1/8! < -1/252 - (1/252 - 1/251) = -1/251.$$
Therefore for all $N \ge 8$ $$-1/250 < \frac{19}{7} - e_{N} < -1/251$$