Disk of convergence of the series $ \sum\limits_{n=1}^\infty n!\,(z-i)^{n!} $

Question

$$ \sum_{n=1}^\infty n!(z-i)^{n!} $$

Find the disk of convergence of this powerseries.

Can I set $n!=k$ and then deal with $\sum_{n=1}^\infty k z^k$ .

On another note $\frac{z^{(n+1)!}}{z^{n!}}$ converges when $|z|<1$.

Answer 1

One knows that $nw^n\to0$ if and only if $|w|\lt1$ hence $n!\,w^{n!}\to0$ if and only if $|w|\lt1$ (a small argument is necessary for the "and only if" direction but you should be able to find it) thus the disk of convergence is centered at $___$ with radius $____$.

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Recall that the disk of convergence $D$ of a series $\sum\limits_na_nw^n$ is characterized by the following pair of properties:

• For every $w$ in the interior of $D$, $a_nw^n\to0$
• For every $w$ not in the closure of $D$, the sequence $(a_nw^n)$ is unbounded

Exercise: Replace the computational solutions of most MSE questions on the subject by a solution based on this "geometrical" characterization and watch the simplifications this shift of emphasis yields.

Answer 2

Almost. You need (a lot of) zero coefficients. Calling the function $f$, we have

$$f(z) = \sum_{k=0}^\infty a_k\cdot z^k,$$

where

$$a_k = \begin{cases} k &, k = n!\\ 0 &, \bigl(\forall n\bigr)(k \neq n!).\end{cases}$$

Now the Cauchy-Hadamard formula gives you the radius of convergence easily.