$\sum_{n=1}^{+\infty} \frac{n^{n-1}v^n}{n!}$ for what value of $v$ this series will be convergent? How to proceed for it?

Question

I am interested in the convergence of the series $$\sum_{n=1}^{\infty} \left( \frac{n^{n-1}}{n!}v^{n} \right).$$ This series defines the tree function.

Answer 1

Since $n! \sim \sqrt{2\pi n} \left(\frac{n}{e} \right)^n$, the convergence will be the same as $\sum \frac{e^n}{n\sqrt{2\pi n}} v^n$. That is, we'll have convergence for $|v|\leq 1/e$.

Answer 2

Hint:

$$|v| \leq \lim_{n \rightarrow \infty}{\left(\frac{(n+1)!}{n!}\frac{n^{n - 1}}{(n + 1)^{(n + 1) - 1}}\right)},$$

in order for the series to be convergent.

Answer 3

As you mentionned $$\sum_{n=1}^{\infty} \left( \frac{n^{n-1}}{n!}v^{n} \right)=-W(-v)$$ and you know that Euler and Lambert established that $W(x)$ is only defined for $x \geq - \frac{1}{e}$.