Are the sums $\sum_{n=1}^{\infty} \frac{1}{(n!)^k}$ transcendental?

Solution 1:

Suppose we have an irreducible polynomial $p(X)=a_dX^d+\ldots+a_1X+a_0\in\Bbb Z[X]$ with $p(\alpha)=0$ and $a_d\ne0$.

For each $N$, we can combine the first $N$ summands and find an integer $A_N$ such that $$\left|\alpha-\frac{A_N}{(N!)^k}\right| =\sum_{n>N}\frac 1{(n!)^k}<\frac 2{((N+1))!^k}.$$ For $x$ close enough to $\alpha$, we have $|p(x)|\le 2|a_d|\cdot|x-\alpha|^d$, hence for $N\gg 0$ and by the MWT, $$\left|p\left(\frac {A_n}{(N!)^k}\right)\right|\le\frac{2^{d+1}|a_d|}{((N+1)!)^{kd}}.$$ On the other hand, $$ (N!)^{kd}p\left(\frac {A_n}{(N!)^k}\right)$$ is a non-zero integer, hence $\ge1$ or $\le -1$. We conclude $$ \frac1{(N!)^{kd}}\le \frac{2^{d+1}|a_d|}{((N+1)!)^{kd}},$$ or $$ (N+1)^{k}\le2\sqrt[d]{|2a_d|}.$$ As this inequality cannot hold for infinitely many $N$, we arrive at a contradiction. We conclude that $p$ as above does not exist and so $\alpha$ is transcendental.