For $n>2$, $\sum_{k=2}^n\frac{1}{k!}$ has a non-terminating decimal representation.

Question

For a given $n>2$, is it true that $\sum_{k=2}^n\frac{1}{k!}$ has a non-terminating decimal representation? First $$\sum_{k=2}^n\frac{1}{k!}=\frac{1+k+k(k-1)+\cdots +k(k-1)\cdots3}{k!}.$$ If $\sum_{k=2}^n\frac{1}{k!}$ has a terminating decimal representation, $$\frac{1+k+k(k-1)+\cdots +k(k-1)\cdots3}{k!}=\frac{a}{10^m}$$ for some $m$ and $a<10^m$. I don't know what to do next or if $\sum_{k=2}^n\frac{1}{k!}$ has a terminating decimal representation for some $n$.

Answer 1

$$\sum_{k=2}^n\frac{1}{k!}=\frac{1\!+\!n\!+\!n(n\!-\!1)\!+\!n(n\!-\!1)(n\!-\!2)\!+\!\ldots\!+\!n(n-1)\!\ldots\!3}{n!}$$

cannot have a terminating decimal representation for $\,n>2\,$ because $\,3\,$ divides $\,n!\,$ but $\,3\,$ does not divide

$1+n+n(n-1)+\ldots+n(n-1)\ldots3\;,$

indeed $\,3\,$ does not divide $\,\begin{cases}1+n\quad&\text{if }n=3\\1\!+\!n\!+\!n(n\!-\!1)\!=\!1\!+\!n^2&\text{if }n>3\end{cases}$

but $\,3\,$ divides $\;n(n\!-\!1)(n\!-\!2)\!+\!\ldots\!+\!n(n\!-\!1)\!\ldots\!3\;\;$ if $\;n>4$

since it is a sum of products of three or more consecutive positive integers.

Addendum :

$3\,$ does not divide $\,1+n^2\;\;$ because

$n^2\equiv0\!\!\!\!\mod\!\!3\;$ or $\;n^2\equiv1\!\!\!\!\mod\!\!3\;,\;\;$ hence ,

$1+n^2\equiv1\!\!\!\!\mod\!\!3\;$ or $\;1+n^2\equiv2\!\!\!\!\mod\!\!3\;.$