= Solution
The key <telescoping series> identity is
$$
c_k=\frac{(q+k-1)q!}{(q+k)!}=\frac{q!}{(q+k-1)!}-\frac{q!}{(q+k)!}.
$$
For the requested <mathematical induction>, when $n=1$ the proposed sum is $q/(q+1)=1-q!/(q+1)!$. If the identity holds for $n$, adding $c_{n+1}=q!/(q+n)!-q!/(q+n+1)!$ cancels its last term and yields $1-q!/(q+n+1)!$. Thus it holds for every $n\geq1$. Since $q!/(q+n)!\to0$ for fixed positive <integer> $q$, the infinite <series> has sum $\boxed{1}$.
For the second <series>, the integral digit condition is $0\leq a_n\leq n-1$, and in particular $a_1=0$. Its partial sums are increasing and satisfy
$$
0\leq\sum_{n=1}^N\frac{a_n}{n!}\leq\sum_{n=1}^N\frac{n-1}{n!}=1-\frac1{N!}\leq1.
$$
Thus the <bounded monotone sequence theorem> proves convergence to a <real number> $S\in[0,1]$. The same telescoping calculation gives, for every $N\geq1$,
$$
\sum_{n>N}\frac{n-1}{n!}=\frac1{N!},\qquad R_N:=N!\left(S-\sum_{n=1}^N\frac{a_n}{n!}\right)\in[0,1].
$$
If positive digits occur infinitely often, then every tail contains a positive term, so $R_N>0$. If $a_n\leq n-2$ occurs infinitely often, then every tail contains a digit smaller than its maximum. Indeed,
$$
1-R_N=N!\sum_{n>N}\frac{n-1-a_n}{n!}>0.
$$
Consequently the two conditions together give $0<R_N<1$ for every $N$. Were $S=A/B$ a <rational number>, with positive <integer> $B$, choose $N\geq B$. Because $B\mid N!$, the number $N!S$ is an <integer>, as is $\sum_{n=1}^Na_nN!/n!$. Their difference $R_N$ would be an <integer> strictly between zero and one, a contradiction.
Conversely, if positive digits occur only finitely often, then $S$ is a finite sum of <rational numbers>. If digits below their maximum occur only finitely often, then $a_n=n-1$ for every $n>N$ for some $N$, giving
$$
S=\sum_{n=1}^N\frac{a_n}{n!}+\frac1{N!}\in\mathbb Q.
$$
These are exactly the two ways the conjunction of infinitude conditions can fail. Hence the <irrationality criterion for factorial series> is proved in both directions: \b[$S$ is irrational exactly when both kinds of digit occur infinitely often.]
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