Taylor coefficients of $\exp⁡(1/(1-z))$

Question

I'm specifically interested in estimating the growth of the coefficients $(a_n)_{n\in\mathbb{N}}$ of the Taylor series of $$\exp⁡(1/(1-z))$$ centered in $0$: knowing that $\limsup_{n\rightarrow+\infty}|a_n|^{1/n}=1$, what I'm trying to figure out is whether the growth of $(a_n)_{n\in\mathbb{N}}$ is at most polynomial or super-polynomial... I tried to calculate explicitly the coefficients in order to find some regularity in the sequence of the coefficients, but I quickly got lost in the process.

Thanks in advance for any answer or suggestion.

Answer 1

Here's a quick proof of the leading asymptotic term. Use the associated Laguerre polynomial generating function (Gradshteyn 8.975.1) to determine that $$e^{-1}e^{1/(1-z)} = e^{z/(1-z)} = \sum_{n=0}^\infty L_n^{-1}(-1)z^n.$$ Use a contour integral representation of the Laguerre polynomials to show that $$L_n^{-1}(-1)=\frac{1}{n\,2\pi\,i} \oint (1+1/t)^n\,e^t\,dt.$$ The contour will be deformed through the saddle point which occurs at $t_0=\tfrac{1}{2}(-1+\sqrt{4n+1}).$ Saddle point analysis yields the approximation $$L_n^{-1}(-1) \sim \exp{(2\sqrt{n} -1/2)}\, \frac{1}{n\,2\pi\,i}\lim_{a \to \infty}\int_{-ia}^{ia} \exp{(c(t-t_o)^2)}\,dt.$$ In this formula the leading exponential is the asymptotic expansion for $n \to \infty$ of $(1+1/t_0)^n\,e^{t_0}.$ The $c$ in the equation, again keeping only the largest term, is $c=1/\sqrt{n}.$A rigorous application of saddle point method will show that value of all but that on a vertical line segment can be neglected, and that segment can be extended to $\infty$ with negligible affect, and the integral becomes the well-known Gaussian. Collecting results we find that $$L_n^{-1}(-1) \sim \frac{\exp{(2\sqrt{n} -1/2)}}{ 2\sqrt{\pi} \, n^{3/4} } $$ It is interesting that a naive first attempt argued as follows: $$L_n^{-1}(-1) = \frac{1}{n} \sum_{m=1}^n \binom{n}{m} \frac{m}{m!} \sim \sum_{m=1}^n \frac{n^m}{m!} \frac{m}{m!} \sim \sum_{m=1}^\infty \frac{n^m}{m!} \frac{m}{m!}=\frac{1}{\sqrt{n}}\,I_1(2\sqrt{n}).$$ One can use the known asymptotics of the Bessel function and get an expansion that agrees with the above $\textit{except for a factor of }$ $\exp{(-1/2)}.$ I thought that $\binom{n}{m} \sim n^m/m!$ would certainly work since the weight of $1/(m-1)!$ heavily emphasizes the beginning terms of the sum. Numerical evidence shows that it wasn't. I should've remembered that the approximation assumes fixed $m$, but there is a summation so $m$ can get quite large. It's interesting that a mistake only ended up with a different constant before the asymptotic expression.

Answer 2

A simple calculation (see this question) yields$$ \exp \left(\frac{1}{1-z}\right) = \sum \limits_{n=0}^\infty \left(\frac{1}{n!} \sum \limits_{k=0}^\infty \frac{(k)_n}{k!}\right) z^n $$ with the Pochhammer symbol $(k)_n = k (k+1) \cdots (k + n - 1)$ .

Mathematica manages to evaluate the sum over $k$ in terms of the confluent hypergeometric function: $$ a_n = \frac{1}{n!} \sum \limits_{k=0}^\infty \frac{(k)_n}{k!} = {_1}F_1 (n+1;2;1) = \sum \limits_{k=0}^\infty \frac{(n+1)_k}{(2)_k k!} \, , \, n \in \mathbb{N}_0 \, . $$ I have no idea how to derive this result, since the partial sums do not even agree, but it seems accurate numerically.

The asymptotic behaviour of the hypergeometric function is documented here (equation 13.8.12; $_1 F_1$ is called $M$) and the leading order term yields $$ a_n \sim \sqrt{\frac{\mathrm{e}}{4 \pi}} \frac{\mathrm{e}^{2 \sqrt{n}}}{n^{3/4}} \, , n \rightarrow \infty \, ,$$ so the growth is indeed superpolynomial.