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What is the radius of convergence of $\sum_{n=0}^{\infty} z^{n!}$?

I tried using ratio test and root test , applying the latter leaves me with the same type of problem again that is with $\sum_{n=0}^{\infty} z^{(n-1)!}$ ?.

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    Hint: If $|z|<1$ then $|z^{n!}|<|z^n|$. If $|z|\geq1$ then $|z^{n!}|\geq1$. – 2017-02-05
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    While the ratio test and root test are important tools, @YuxiaoXie gives a simple, straightforward argument which is very useful for understanding the series. – 2017-02-05

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Let's check that ratio test:

$$\left|\frac{z^{(n+1)!}}{z^{n!}}\right|=|z^{(n+1)!-n!}|=|z^{n(n!)}|$$

Let $n(n!)=k$ so that we have

$$\lim_{n\to\infty}\left|\frac{z^{(n+1)!}}{z^{n!}}\right|=\lim_{k\to\infty}|z^k|$$

Can you take it from there?

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    Yup..ok..through ratio test. – 2017-02-05
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It looks like you misunderstood the root test.
You say you get a sum when applying it, but you shouldn't get that.

We have $a_n=z^{n!}$. Therefore $\sqrt[n]{a_n} = \sqrt[n]{|z^{n!}|} = |z^{(n-1)!}|$.

Now, the root test states that if $\lim_{n\to\infty} \sqrt[n]{|a_n|} <1$, the series converges, if $\lim_{n\to\infty} \sqrt[n]{|a_n|} >1$ the series diverges. So you want to find $\lim_{n\to\infty}|z^{(n-1)!}|$. For which $z$ is this larger than one and for which $z$ is this smaller than one?

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    for $|z| < 1$ the series should converge right and for $|z|>1$ the series should diverge – 2017-02-05
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    @BAYMAX Yup, that's the radius of convergence. – 2017-02-05
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    @BAYMAX Yes. You are missing one case, and that is $|z|=1$. This can't be handled by the root test or the ratio test, but should be straightforward. – 2017-02-05
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    If I may check up with you, it diverges for all $|z|=1$, right? – 2017-02-05
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    @SimplyBeautifulArt Yes, because $z=e^{i\theta}$. If $\theta = \frac{p}{q}$, then $z^{n!}=e^{i\theta n!} = 1$ for large enough $n$. If $\theta$ isn't rational, it follows by continuity. – 2017-02-05
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Hadamard's formula gives the answer at once:

If $\sum_n a_n zn$ is a power series, its radius of convergence $R$ is given by $$\frac1R=\limsup_{n\to\infty} \bigl(\lvert a_n\rvert^{1/n}\bigr)=1.$$