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

This question is inspired by my answer to the question "How to compute $\prod_{n=1}^\infty\left(1+\frac{1}{n!}\right)$?".

The sums $f(k) = \sum_{n=1}^{\infty} \frac{1}{(n!)^k}$ (for positive integer $k$) came up, and I noticed that $f(1) = e-1$ was transcendental and $f(2) = I_0(2)-1$ (modified Bessel function) was probably transcendental since $J_0(1)$ (Bessel function) is transcendental.

So, I made the conjecture that $all$ the $f(k)$ are transcendental, and I am here presenting it as a question.

The only progress I have made is to show that all the $f(k)$ are irrational.

This follows the standard proof that $e$ is irrational: if $f(k) = \frac{a}{b} =\sum_{n=1}^{\infty} \frac{1}{(n!)^k}$, multiplying by $(b!)^k$ gives $a (b!)^{k-1}(b-1)! =\sum_{n=1}^{b} \frac{(b!)^k}{(n!)^k} +\sum_{n=b+1}^{\infty} \frac{(b!)^k}{(n!)^k}$ and the left side is an integer and the right side is an integer plus a proper fraction (easily proved).

I have not been able to prove anything more, but it somehow seems to me that it should be possible to prove that $f(k)$ is not the root of a polynomial of degree $\le k$.

• By the way, the correct way to produce italics is *italics*. See here for a guide to formatting with Markdown. Aug 7, 2013 at 2:08
• But $italics$ is $so$ much easier. Aug 7, 2013 at 2:56
• Perhaps you could have a look at Roth's Theorem (1955, Fields Medal). If you show that the inequality is not true, then you will have proved that it's transcendental. Aug 13, 2013 at 13:04
• Or perhaps just the Liouville theorem. Any number that has an infinity of "good" rational approximations can not be algebraic. One standard example is $\sum_{n=1}^{\infty}10^{-n^2}$, and the example in the question fits this scheme. Feb 13, 2014 at 19:09
• @LutzL I can't say much about the larger values of $k$, but for $k=1$ the continued fraction of $e$ is well-known and its coefficients are nowhere near large enough to prove transcendence by the Roth bound, let alone the weaker Liouville bound. Which example are you referring to that fits the scheme? I believe the set of numbers which violate the conclusion of Roth's theorem is measure zero, so generally one does not expect a typical transcendental number to be amenable to this method without a specific reason to believe so. May 19, 2016 at 20:13

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.
• I'm a little suspicious of this answer, because it seems too simple. For example, if $k=1$, it offers an extremely short proof that $e$ is transcendental. I'll look at that case more closely. Jan 13, 2017 at 19:21
• I think I see a problem. You say that "For $x$ close enough to $\alpha$, we have $|p(x)|\le 2|a_d|\cdot|x-\alpha|^d$". This implies that $\alpha$ is a $d$-fold root. However, if $p'(\alpha) \ne 0$, all that can be said is that $|p(x)| = O(|x-\alpha|)$. Jan 13, 2017 at 19:34
• @Mefitico This wwas meant to stand for Mean Value Theorem. Howeverm now that I notice marty chohen's comment, the proof seems to work only for $d=1$. In other words, I think this only shows that the numbers are irrational, not that they are transcendental ... Nov 9, 2018 at 22:38