N!be might not be an integer

b is an integer. For some $$n \geq 0$$, $$n!be$$ is not an integer.

Explanation

$$e = 1 + \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + \cdots $$ (*)

$$n!be = n!b\left(1 + \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + \cdots + \frac{1}{n!} \right) + n!b\left(\frac{1}{(n+1)!} + \frac{1}{(n+2)!} + \cdots \right) $$

The first part $$n!b\left(1 + \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + \cdots + \frac{1}{n!} \right)$$ is always an integer.

The second part $$n!b\left(\frac{1}{(n+1)!} + \frac{1}{(n+2)!} + \cdots \right) $$ is not an integer, for some value of n (more)