= Solution
Let
$$
S=\sum_{n=1}^{\infty}\frac{m_n}{n!}.
$$
Assume for a contradiction that $S=a/b$ is <rational number>[rational]. For every $N\geq b$, both $N!S$ and
$$
N!\sum_{n=1}^{N}\frac{m_n}{n!}
$$
are <integers>, because the $m_n$ are integers. Their positive difference
$$
R_N=N!\sum_{n=N+1}^{\infty}\frac{m_n}{n!}
=\sum_{k=1}^{\infty}
\frac{m_{N+k}}{(N+1)\cdots(N+k)}
$$
must therefore be a positive integer.
On the other hand, <concavity> of $x^\alpha$ for $0\leq\alpha<1$ gives $(N+k)^\alpha\leq N^\alpha+k^\alpha\leq N^\alpha+k$. Since $(N+1)\cdots(N+k)\geq(N+1)^k$,
$$
0<R_N
\leq A\sum_{k=1}^{\infty}\frac{N^\alpha+k}{(N+1)^k}
=A\left(N^{\alpha-1}+\frac{N+1}{N^2}\right)
\longrightarrow0.
$$
Eventually $0<R_N<1$, contradicting its integrality. Hence
$$
\boxed{\sum_{n=1}^{\infty}\frac{m_n}{n!}\text{ is irrational}.}
$$
The integrality assumption is essential. For example, take $\alpha=0$, $A=2$, and
$$
m_n=\frac2{e-1}\qquad(n\geq1).
$$
Then $1<m_n<2$, but the <exponential series> gives
$$
\sum_{n=1}^{\infty}\frac{m_n}{n!}
=\frac2{e-1}\sum_{n=1}^{\infty}\frac1{n!}=2.
$$
Thus \b[the result does not remain true for arbitrary real $m_n$].
Back to article page
