prove that $\displaystyle \Sigma_{j=1} ^\infty 2^{-2^j} z^{-j}$ converges for $\displaystyle z \neq 0$ and defines a function which has an essential singularity at P=0.
The series...
$\displaystyle \varphi(z)= \sum_{j=1}^{\infty} \frac{1}{(4 z)^{j}}$ (1)
... is a Laurent series with an infinite numers of negative powers of z, so that in $\displaystyle z=0$ is has an essential singularity. If we consider (1) as a 'geometric series' we have...
$\displaystyle \varphi(z) = \frac{\frac{1}{4z}}{1 - \frac{1}{4z}} = \frac{1}{4 z-1}$ (2)
... and the (1) converges for $\displaystyle |z|>\frac{1}{4}$ , not symply for $\displaystyle z\ne 0$ ...
Kind regards
$\displaystyle \chi$ $\displaystyle \sigma$
Very sorry for my misreading! ...
According to the 'Stirling approximation' is...
$\displaystyle \ln \frac{1}{j!} \approx -j\cdot \ln j + j > -2^{j}\cdot \ln 2 = \ln 2^{-2^{j}}, \forall j>0$ (1)
... so that is...
$\displaystyle \frac{1}{j!} > 2^{-2^{j}}, \forall j>0$ (2)
Now the series $\displaystyle \sum_{j=1}^{\infty} \frac{1}{j!\cdot z^{j}}$ absolutely converges $\displaystyle \forall z\ne 0$ and therefore the same is for the series $\displaystyle \sum_{j=1}^{\infty} \frac{2^{-2^{j}}}{z^{j}}$...
Kind regards
$\displaystyle \chi$ $\displaystyle \sigma$