38. If $h=\sum_{n=1}^{+\infty} \frac{\sigma(n)}{n!}=\frac{r}{s}$ is a rational number, where $(r, s)=1$, and let $p>\max \{s, 6\}$ be a prime number. From $h=\sum_{n=1}^{p-1} \frac{\sigma(n)}{n!}+\sum_{n=p}^{+\infty} \frac{\sigma(n)}{n!}$, we get
$(p-1)!h=(p-1)!\sum_{n=1}^{p-1} \frac{\sigma(n)}{n!}+\sum_{k=0}^{+\infty} \frac{\sigma(p+k)}{p(p+1) \cdot \cdots \cdot(p+k)}$.
Let $m=\sum_{k=0}^{+\infty} \frac{\sigma(p+k)}{p(p+1) \cdot \cdots \cdot(p+k)}$, since
$\sigma(p)=p+1, \sigma(p+k)6$, thus $1<m<1+\frac{1}{p}+\frac{p-1}{p}=2$. Since $(p-1)!h$ and $(p-1)! - \sum_{n=1}^{p-1} \frac{\sigma(n)}{n!}$ are both integers, and $m$ is not an integer, this leads to a contradiction. This proves that $\sum_{n=1}^{+\infty} \frac{\sigma(n)}{n!}$ is an irrational number.