$S = \frac{1}{q+1}+\frac{1}{(q+2)(q+1)}+\frac{1}{(q+3)(q+2)(q+1)}...$
I know that $0<S<1$. But is it rational?
I took this series from(proof by contradiction that $e$ is is irrational ): http://www.mathshelper.co.uk/Proof%20That%20e%20Is%20Irrational.pdf
In the paper it assumes that $e=p/q$. So $q$ is not allowed to be $0$.
On
\begin{eqnarray*} \underbrace{e}_{\text{irrational}}=\underbrace{1+\frac{1}{1!}+ \cdots+ \frac{1}{q!}}_{rational} +\frac{1}{q!}\left( \frac{1}{q+1} +\frac{1}{(q+1)(q+2)}+\cdots \right) \end{eqnarray*} so ...
On
If this sum be rational, then according to the paper you linked above, the number $e$ will be a rational number.
On
I doubt $S(q)$ be rational; in particular $S(0)=e-1$ which is trascendental. Actually $$S(q)=\begin{eqnarray*} e=1+\frac{1}{1!}+ \cdots+ \frac{1}{q!} +\frac{1}{q!}\left( \frac{1}{q+1} +\frac{1}{(q+1)(q+2)}+\cdots \right) \end{eqnarray*}$$ Consequently $S(q)$ is always trascendental because of it is equal to $e$ minus a rational.
It is irrational in general for positive integer $q$.
This sum represents the Engel expansion of some number. Engel expansion is unique, and it is finite if and only if the number is rational.
(If every $a_k$ is different though!)
Since in this case the expansion is infinite by definition, the number is irrational.