Given integers $a_1,a_2,…$ such that $1\le a_n\le n-1$ , $n=2,3,…$
Show that the sum of the series $\underset{n=1}{\overset{\infty}{\sum}}\frac{a_n}{n!}$ is rational if and only if there exists integer $N$ such that $a_n=n-1$ for all $n\ge N$.
My attempt
If part: $a_n=n-1$ $\forall n\ge N$
$\Rightarrow\frac{a_n}{n!}=\frac{n-1}{n!}=\frac{1}{(n-1)!}-\frac{1}{n!}$ $\forall n\ge N$
$\Rightarrow\underset{k=N}{\overset{n}{\sum}}\frac{a_k}{k!}=\frac{1}{(N-1)!}-\frac{1}{n!}$
$\Rightarrow\underset{n=1}{\overset{\infty}{\sum}}\frac{a_n}{n!}=\underset{n\rightarrow\infty}{\lim}\underset{n=N}{\overset{n}{\sum}}\frac{a_n}{n!}=\frac{1}{(N-1)!}\in\mathbb{Q}$
I need help with the only if part.
Best Answer
Suppose that
$$\sum_{n=1}^\infty\frac{a_n}{n!}=\frac{b}{c}.$$
Multiply both sides by $c!$ to get
$$\sum_{n=1}^cc!\frac{a_n}{n!}+\sum_{n=c+1}^\infty c!\frac{a_n}{n!}=b(c-1)!.$$
Now, note that the first term on the left and the term on the right are both integers. In addition, since
$$\sum_{n=c+1}^\infty c!\frac{a_n}{n!}>0,$$
we have
$$\sum_{n=c+1}^\infty c!\frac{a_n}{n!}\in\mathbb{N}$$
(note here that $\mathbb{N}=\{1,2,3,\cdots\}$). Now, if not all $a_n=n-1$, we have that
$$\sum_{n=c+1}^\infty c!\frac{a_n}{n!}<\sum_{n=c+1}^\infty c!\frac{n-1}{n!}$$
(note the strictly less than condition). However, this is
$$\sum_{n=c+1}^\infty c!\frac{a_n}{n!}<\sum_{n=c+1}^\infty c!\frac{n-1}{n!}=1.$$
Thus, $\sum_{n=c+1}^\infty c!\frac{a_n}{n!}$ is an integer between $0$ and $1$, an impossibility. We conclude
$$\sum_{n=1}^\infty \frac{a_n}{n!}\not\in\mathbb{Q}.$$