Series $z^{n!}$ and the geometric series

Question

For the geometric series $$ \sum_{n=0}^\infty z^n =\lim_{n \rightarrow \infty} \frac{1}{1-z^n} $$ for $$ |z|<1 $$

How does the series $$ \sum_{n=0}^\infty z^{n!} $$ compare?

$$ \sum_{n=0}^\infty z^{n!} = \sum_{n=0}^\infty z^{n(n-1)!} = 1 + z + z^2 + z^6 + z^{24} + \dots + z^\infty $$ so it should react similarly since it contains some elements of $z^n.$

Is the claim that infinity allows for the same quotient to be used true?

such that:

$$ \sum_{n=0}^\infty z^n = \sum_{n=0}^\infty z^{n!} = \lim_{n \rightarrow \infty} \frac{1}{1-z^{n!}} $$ for $$ |z|<1 $$

Answer 1

\begin{align} \require{cancel} \sum_{n=0}^\infty z^n = \xcancel{\frac{1}{1-z^n}} \\[8pt] \updownarrow\quad \\[8pt] \sum_{n=0}^\infty z^n = \frac{1}{1-z} \end{align} Note: $$ \sum_{n=0}^\infty z^n = \underbrace{1 + z + z^2 + z^3 + z^4 + \cdots}_\text{No “$n$'' appears here.} $$ The variable $n$ is bound, not free. The expression $\text{“}\sum_{n=0}^\infty\text{''}$ works as a variable-binding device. The variable $n$ is different in different terms on the right side of the above equality. There is no $\text{“}n\text{''}$ outside that context. In the expression $\dfrac 1 {1-z^n}$ the variable $n$ is free. What number would $n$ be in that expression? So that expression cannot have the same value as the sum you started with.

The other series, however, has the same radius of convergence.

Answer 2

Note $$ \sum_{n=0}^\infty|z|^{n!}\neq \sum_{n=0}^\infty|z|^{n}. $$ For $|z|<1$, one has $$ |z|^{n!}\le|z|^n $$ since $n!\ge n$. So $$ \sum_{n=0}^\infty|z|^{n!}\le \sum_{n=0}^\infty|z|^{n}=\frac{1}{1-|z|} $$ and hence $\sum_{n=0}^\infty|z|^{n!}$ converges absolutely for $|z|<1$.