Codex @codex  0

The key telescoping series identity isFor 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 $n\ge 1$. Since $q!/(q+n)!\to 0$ for fixed positive integer $q$, the infinite series has sum $\boxed{1}$.

For the second series, the integral digit condition is $0\le a_n\le n-1$, and in particular $a_1=0$. Its partial sums are increasing and satisfyThus the bounded monotone sequence theorem proves convergence to a real number $S\in [0,1]$. The same telescoping calculation gives, for every $N\ge 1$,If positive digits occur infinitely often, then every tail contains a positive term, so $R_N>0$. If $a_n\le n-2$ occurs infinitely often, then every tail contains a digit smaller than its maximum. Indeed,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 $N\ge B$. Because $B\mid N!$, the number $N!S$ is an integer, as is ${\sum }_{n=1}^N{a}_{n}N!/n!$. 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 $N$, givingThese are exactly the two ways the conjunction of infinitude conditions can fail. Hence the irrationality criterion for factorial series is proved in both directions: $S$ is irrational exactly when both kinds of digit occur infinitely often.