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I am working in a field not really based on combinatorics, therefore I appologize if my question is in any kind invalid. Nevertheless, in my calculations, the Bell numbers appeared. I need to find some $x$ such that the ordinary generating function $$B(x) = \sum_{n=0}^{\infty}B_n x^n$$ converge. I haven't found the answer nowhere in the literature. On the opposite, there are quite a lot of results concerning $B(x)$, but none of them is questioning for which $x$ it has some sense. It is evident that the case $x>1$ lead to a divergent series, which is not much of an interest. But what about $x<1$? I suppose there must be such $x$, otherwise it is nonsense to study such series, is it not?

One other thing suprised me. There is a nice representation in Klazar of $B(x)$ such that $$B(x) = \sum_{n=0}^{\infty}\frac{x^n}{(1-x)(1-2x)\cdots(1-n x)}.$$ But what if $x$ equals to $1/k$ for some $k$ natural? Would that make the series divergent? I am sorry, but is has been bugging me for some time that there is no explanation in the literature that I have been looking into. Does anyone have some relevant source of information?

Thank you, I would appreciate any help.

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  • $\begingroup$ Are you referring to this article by Klazar? Because its Proposition 3.4 does make a (negative) claim about convergence, although maybe not the one you're looking for. My complex analysis has gotten really rusty, but I thought if a power series converges at some $z = z_0 \in \mathbb C$, then it should be analytic in the open ball of radius $\left|z_0\right|$ around $0$; is that true? If so, then I think it rules out convergence anywhere other than at $0$. $\endgroup$ Commented Jan 5, 2021 at 20:11
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    $\begingroup$ Actually, the asymptotic expression for $\dfrac{\ln B_n}{n}$ ascribed to de Bruijn in en.wikipedia.org/wiki/Bell_number#Growth_rate should also preclude convergence of the series anywhere other than at $0$. $\endgroup$ Commented Jan 5, 2021 at 20:18
  • $\begingroup$ Thank you, that actually helped me! $\endgroup$
    – Daniela
    Commented Jan 5, 2021 at 20:22
  • $\begingroup$ Title of this article referenced by @darijgrinberg: Klazar - Bell numbers, their relatives, and algebraic differential equations. $\endgroup$
    – LSpice
    Commented Jan 5, 2021 at 20:37

1 Answer 1

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The Bell numbers satisfy $\frac{\ln B_n}{n} \sim \ln n$ which is faster than exponential, so the ordinary generating function $\sum B_n x^n$ has zero radius of convergence. As a more elementary argument, Dobinski's formula

$$B_n = \frac{1}{e} \sum_{k \ge 0} \frac{k^n}{k!}$$

establishes that $B_n$ grows at least as fast as $k^n$ for any positive integer $k$, which also implies that $B_n$ grows faster than exponentially and so $\sum B_n x^n$ has zero radius of convergence.

This does not imply that it's nonsense to study this series; formal power series can be studied abstractly and it's common practice to do so in combinatorics. $x$ is just never specialized to a concrete value in $\mathbb{C}$ and we never perform operations that would involve adding infinitely many coefficients. As another example, $\sum n! x^n$ makes sense as a formal power series also despite having zero radius of convergence, and there are various interesting things to say about it, e.g. its logarithm

$$\log \sum n! x^n = x + \frac{3}{2} x^2 + \frac{13}{3} x^3 + \frac{71}{4} x^4 + \dots $$

counts subgroups of the free group $F_2$. There is also a funny continued fraction

$$\sum n! x^n = \frac{1}{1 - x - \frac{x^2}{1 - 3x - \frac{4x^2}{1 - 5x - \frac{9x^2}{1 - 7x - \frac{16x^2}{1 - 9x - \dots}}}}}$$

coming from the fact that $n!$ is the sequence of moments of the exponential distribution. $B_n$ is the sequence of moments of the Poisson distribution with $\lambda = 1$ (this is equivalent to Dobinski's formula) which also gives a continued fraction expansion for $\sum B_n x^n$, namely the second expansion given by Ira Gessel in the comments.

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    $\begingroup$ We also have continued fractions for $B(x) = \sum_{n=0}^\infty B_n x^n$: $$B(x) = \cfrac{1}{1- \cfrac{x}{1- \cfrac{x}{1- \cfrac{x}{1- \cfrac{2x}{1- \cfrac{x}{1- \cfrac{3x}{1- \cfrac{x}{1- \cfrac{4x}{1-\cdots }}}}}}}}} $$ and $$B(x) = \cfrac{1}{1-x- \cfrac{x^2}{1-2x- \cfrac{2x^2}{1-3x- \cfrac{3x^2}{1-\cdots }}}} $$ $\endgroup$
    – Ira Gessel
    Commented Jan 5, 2021 at 21:11
  • $\begingroup$ @Ira: yes! The second one should follow from the fact that $B_n$ is the sequence of moments of the Poisson distribution (with $\lambda = 1$). I got nothing on the first one though. $\endgroup$ Commented Jan 5, 2021 at 21:24

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