calculate the annulus of convergence of $\displaystyle \Sigma _{j=-\infty, j \neq 0} ^\infty z^j / j^j$
The critical point is relative to the sum of the powers of z with negative exponent...
$\displaystyle \sum_{j=1}^{\infty} \frac{(-j)^{j}}{z^{j}}$ (1)
In fact the (1) diverges for any value of $\displaystyle z\ne 0$...
Kind regards
$\displaystyle \chi$ $\displaystyle \sigma$