I'm reading about the asymptotic expansion of the Lambert W function:
$$W(x)=\frac{1^0}{1!}x^1-\frac{2^1}{2!}x^2+\frac{3^2}{3!}x^3-\frac{4^3}{4!}x^4+\dots$$
And it says that the convergence radius of this expansion is $\frac1e$ (according to the ratio test).
I've been playing with a tweak of this expansion, namely $-W(-x)$, which yields all positive terms:
$$-W(-x)=\frac{1^0}{1!}x^1+\frac{2^1}{2!}x^2+\frac{3^2}{3!}x^3+\frac{4^3}{4!}x^4+\dots$$
Is there a way for me to ascertain that the convergence radius of this expansion remains $\frac1e$?
In other words, is it true to say $-\frac1e\leq{x}\leq\frac1e\iff-W(-x)$ converges?
Thank you!
Your series:
$\sum\limits_{n=1}^\infty \frac{n^{n-1}}{n!}x^n$
Ratiotest:
$\lim\limits_{n \to \infty}\frac{(n+1)^{n}n!}{n^{n-1}(n+1)!}=\lim\limits_{n \to \infty}\frac{(n+1)^{n}}{n^{n-1}(n+1)} = \lim\limits_{n \to \infty}\frac{(n+1)^{n-1}}{n^{n-1}}=\lim\limits_{n \to \infty}(\frac{(n+1)}{n})^{n-1}=\lim\limits_{n \to \infty}(1+\frac{1}{n})^{n-1}=\lim\limits_{n \to \infty}(1+\frac{1}{n})^{n}(1+\frac{1}{n})^{-1}$
$\lim\limits_{n \to \infty}(1+\frac{1}{n})^{n}\lim\limits_{n \to \infty}(1+\frac{1}{n})^{-1}=e*1=e \Longrightarrow \,$radius of convergence is $\rho=\frac{1}{e}$