Limiting growth ratio of $[x^n]f(x)^n$

Question

Based on a few examples (mainly data from OEIS as well as a bit of theory) I've arrived at the following conjecture:

Conjecture: Let $f(x)$ be analytic at $0$ and nonlinear, and let $[x^n]f(x)$ represent the coefficient of $x^n$ in its power series. Let $$\alpha:=\lim_{n\to\infty}\frac{[x^{n+1}]\left(f(x)^{n+1}\right)}{[x^n]\left(f(x)^n\right)}$$ assuming it exists. Then $f(x)-\alpha x$ has a double root.

Here's one application of this conjecture. Let $P_n$ be the probability that if you roll an $n$-sided die $n$ times, none of the outcomes appear more than three times. Then using the above conjecture with $f(x)=1+x+\frac{x^2}2+\frac{x^3}6$, together with some generating function know-how, reveals that $$\lim_{n\to\infty}\frac{P_{n+1}}{P_n}=0.97868\ldots=\frac{2.66032\dots}e$$ where $c=2.66032\dots$ is the unique real such that $(1+x+\frac{x^2}2+\frac{x^3}6)-cx$ has a double root. A discriminant calculation reveals that $c$ is algebraic: $8c^3-21c^2-2=0$. (Note to anyone trying to verify this numerically: $P_{n+1}/P_n\cdot(1+\frac1n)^n$ converges faster, and then you can divide the result by $e$.)

Unfortunately, my analytic combinatorics-fu isn't strong enough for me to actually prove this conjecture, so I turn to Stack Exchange. Is this conjecture true, and if so, how can I prove it?

PS. I leave it to the reader to verify that the location of the double root in question is also a root of $xf'(x)-f(x)$. EDIT: In fact, $\alpha$ can also be described as a local extreme value of $f(x)/x$. That might be a simpler way to describe it.

PPS. This is a variation of the fact that if $\beta=\lim_{n\to\infty}[x^{n+1}]f(x)/[x^n]f(x)$ exists then its reciprocal is a pole of $f$, that is, $f(1/\beta)=\infty$. The simplest example of this is $f(x)=1/(1-bx)$ and $[x^n]f(x)=b^n$. This observation is essentially the start of analytic combinatorics, if my understanding is right. EDIT: Actually, $1/\beta$ could be some type of singularity other than a pole; an example would be $f(x)=\sqrt{1-x}$, which has a branch cut at $1/\beta=1$.

Answer 1

Under several additional assumptions, we can prove your conjecture. It's possible that these assumptions can be loosened, but this is not immediately obvious.

Let $g(x)$ be a power series such that $g(0) = 0$ and $g'(0) \neq 0$, and $f(x) = x/g(x)$. By the Lagrange Inversion Theorem $$ g^{-1}(x) = \sum_{n=1}^\infty \frac{x^n}{n} [x^{n-1}] f(x)^n. $$ If the coefficient of a power series are $(c_n)$ then the limit (when it exists) of $|\frac{c_n}{c_{n+1}}|$ is the radius of convergence. Therefore the radius of convergence of $g^{-1}$ is $$ R = \lim_{n\to\infty} \left|\frac{(n+1)[x^{n-1}] f(x)^n}{n[x^n] f(x)^{n+1}}\right| = \lim_{n\to\infty} \left|\frac{[x^{n-1}] f(x)^n}{[x^n] f(x)^{n+1}}\right|. $$ If $[x^n] f(x)^n / [x^{n-1}] f(x)^n = u_n$ and $u_n \sim u_{n+1}$, then $\alpha = 1/R$. If $g^{-1}$ converges on its radius at the point $R$ with an infinite derivative, then $$ g^{-1}(R) = y \implies R = \frac{y}{f(y)} \implies f(y) - \frac{y}{R} = 0 $$ and $$ f'(y) = \frac{g(y) - y g'(y)}{g(y)^2} = \frac{1}{R} - \frac{y g'(g^{-1}(R))}{R^2} = \frac{1}{R} - \frac{y}{R^2 (g^{-1})'(R)} = \frac{1}{R}. $$

Therefore, the function $f(x) - x/R$ has a double root at the point $y$.

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EXAMPLE:

With $f(x) = e^x$, $g(x) = x e^{-x}$ and by the Lagrange inversion theorem, $$ g^{-1}(x) = \sum_{n=1}^\infty \frac{x^n}{n} [x^{n-1}] e^{nx} = \sum_{n=1}^\infty \frac{x^n}{n!} n^{n-1} = -W_0(-x). $$ where $W_0$ is the Lambert W function.

$$ R = \lim_{n\to\infty} \frac{n^{n-1} n!}{(n-1)! (n+1)^n} = \lim_{n\to\infty} \left(1+\frac{1}{n}\right)^{-n} = e^{-1}. $$ We check $$ [x^n] f(x)^n = n^n/n! = n^{n-1}/(n-1)! = [x^{n-1}] f(x)^n $$ therefore $u_n = 1 \sim u_{n+1}$. $$ -W(-e^{-1}) = 1 = \sum_{n=1}^\infty \frac{e^{-n}}{n!} n^{n-1} $$ (the sum converges as it can be compared to a simpler converging series with Stirling's approximation), and the derivative diverges to $\infty$.

Hence $e^x - ex$ is our function with a double root at $1$ (which can be verified easily).