Discuss convergence of the series $\sum \frac{(1+n)^n}{n!}x^n$.

Question

Let $$a_n = \dfrac{(1+n)^n}{n!}$$ Discuss convergence of the series $\displaystyle \sum a_nx^n$.

I've shown that the series converges absolutely when |x| < 1/e and diverged when |x| > 1/e. I can't seem to determine what happens at |x| = 1/e

Answer 1

By Lagrange inversion theorem we have $$ W(x)=\sum_{n\geq 1}\frac{(-n)^{n-1}}{n!} x^n $$ for any $|x|<\frac{1}{e}$, where the Lambert function $W(x)$ is the inverse function of $x\mapsto xe^x$. In particular $$ -W(-x) = \sum_{n\geq 1}\frac{n^{n-1}}{n!} x^n,\qquad -\frac{W(-x)}{x(1+W(-x))}=\sum_{n\geq 1}\frac{n^{n-1}}{(n-1)!}x^{n-1} $$ and by reindexing $$ -\frac{W(-x)}{x(1+W(-x))} = \sum_{n\geq 0}\frac{(n+1)^n}{n!}x^n. $$ Obviously $W\left(-\frac{1}{e}\right)=-1$, hence the RHS is divergent at $x=\frac{1}{e}$. Similarly $W\left(\frac{1}{e}\right)\approx \frac{428}{1537}$ leads to $$ \sum_{n\geq 0}\frac{(n+1)^n (-1)^n}{n! e^n} \approx 0.592.$$