## Loser66 one year ago Find the radius of convergence: $$\sum_{n=0}^{\infty} z^{n!}$$ Please, help

1. anonymous

hold on

2. dan815

-1 to 1

3. Loser66

how?

4. dan815

lol im jk

5. dan815

hmmm greater than -1 and less than 1

6. dan815

7. dan815

|dw:1442626076018:dw|

8. Loser66

$$\sum_{n=0}^\infty z^{n!}= \sum_{n=0}^\infty a_m z^{n!}$$ where $$a_m = \begin{cases}0 ~~if ~~ m \neq n!\\m ~~if ~~m = n!\end{cases}$$

9. Loser66

but I don't know how to deal with the first z, since $$sum_{n=0}^\infty z^{n!}= z^{0!} + z^{1!}+z^{2!}+...... = \color{red}{z} + \sum something we can handle$$

10. Loser66

the red z kills me. hehehe ...

11. Loser66

if n = 1 to infinitive, that is we don't have that red z, then apply Dirichlet's test, we can have the radius of convergence is $$R= \dfrac{1}{lim sup \sqrt[m] {a_m}}=1$$