Why am I getting two different convergence radii for the Taylor series of the Lambert W function at 0? The Taylor series of the Lambert $W$ function at $0$ is given by
$$
W(z) = \sum_{n=1}^\infty \frac{(-n)^{n-1}}{n!}z^n
$$
Now, I read that the convergence radius of the series is $1/e$. And indeed, if I use the ratio test, I can see
$$
\frac{(n+1)^n}{(n+1)!} \frac{n!}{n^{n-1}} = \left(1+\frac{1}{n}\right)^{n-1}\overset{n\to \infty}{\longrightarrow} e
$$
However, I also tried to use the root test and got
$$
\left(\frac{n^{n-1}}{n!}\right)^{\frac{1}{n}} = \exp \left(\ln\left(\frac{n^{n-1}}{n!}\right)\cdot \frac{1}{n}\right)
$$
Focussing on the argument, we have (using $\ln n! \sim n \ln n$)
$$
\ln\left(\frac{n^{n-1}}{n!}\right)\cdot \frac{1}{n} = \frac{(n-1)\ln n - \ln n!}{n} \sim \frac{(n-1)\ln n - n \ln n}{n} = - \frac{\ln n}{n} \overset{n\to \infty}{\longrightarrow} 0
$$
Implying
$$
\left(\frac{n^{n-1}}{n!}\right)^{\frac{1}{n}} \overset{n\to \infty}{\longrightarrow} \exp(0) = 1
$$
I guess my mistake lies in the "$\sim$"-step, however I don't know why this argumentation should be wrong.
I would be very interested in where I made a mistake and also how one can show that the limit of the root is actually $e$.

Yes, the problem lies in the $\sim$ part. Here's another example about how that doesn't work: take $f(x)=x+\frac1x$ and $g(x)=x$. Then $\lim_{x\to\infty}f(x)-g(x)=0$. However,$$\lim_{x\to\infty}\frac{(x-1)f(x)-xg(x)}x=-1.$$
