N!b... might not be an integer

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

Explanation

$$ \frac{1}{n+1} + \frac{1}{(n+1)(n+2)} + \cdots < \frac{1}{n+1} + \frac{1}{(n+1)^2} + \cdots = \frac{b}{n} $$ (more)

For n > b, $$ 0 < n!b\left(\frac{1}{(n+1)!} + \frac{1}{(n+2)!} + \cdots \right) < \frac{b}{n} < 1 $$