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 $\le k$.

marty cohen
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    By the way, the correct way to produce *italics* is `*italics*`. [See here](http://math.stackexchange.com/help/formatting) for a guide to formatting with Markdown. – Zev Chonoles Aug 07 '13 at 02:08
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    But $italics$ is $so$ much easier. – marty cohen Aug 07 '13 at 02:56
  • **`Shift`** **`4`** vs. **`Shift`** **`8`**? (Of course it's also not correct semantically - if you prefer using MathJax, it would be best to use `$\textit{italics}$`). – Zev Chonoles Aug 07 '13 at 03:00
  • How about standard proof of *'e is transcendental'*? Does it help somehow? – Kunnysan Aug 07 '13 at 05:02
  • That had occurred to me, but I haven't had time to refamiliarize myself with the proof to see if it could be generalized. – marty cohen Aug 07 '13 at 05:40
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    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. – user88595 Aug 13 '13 at 13:04
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    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. – Lutz Lehmann Feb 13 '14 at 19:09
  • Related: http://math.stackexchange.com/questions/367183 – Bart Michels Feb 21 '15 at 14:34
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    @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. – Erick Wong May 19 '16 at 20:13

1 Answers1


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.

Hagen von Eitzen
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    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. – marty cohen Jan 13 '17 at 19:21
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    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 $|p(x)| = O(|x-\alpha|)$. – marty cohen Jan 13 '17 at 19:34
  • Could you please clarify in the answer what is meant by "MWT"? I'm guessing it's not "Maximum Weight Trace" nor "minimum-weight spanning-tree" to tell a few colliding acronyms that come to mind. Provide a link if possible. – Mefitico Nov 09 '18 at 16:49
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    @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 ... – Hagen von Eitzen Nov 09 '18 at 22:38