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I am having trouble determining the radius of convergence for the following sums. I have tried ratio test, but havnt gotten anywhere. I don't know what other methods i can apply:

$\sum_{n=0}^{\infty}n!z^{n!}$

$\sum_{n=0}^{\infty}z^{3n}2^n/n!$

What other methods could be used to find the radius of convergence?

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    $\begingroup$ $\lim \sup_{n\to \infty} |a_n|^{1/n}$ ? And why do you think the ratio test doesn't work (the ratio $c_n/c_{n+1}$ where $c_n = n! z^{n!}$) $\endgroup$
    – reuns
    Nov 10, 2016 at 5:27

2 Answers 2

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Let's try the ratio test for the first series. One has

$${a_ {n+1}\over a_n}={(n+1)!\over n!}=n+1\to +\infty$$

And the radius of convergence is $0$. For the second one, we can rewrite it

$$\sum_{n=0}^{+\infty} {\left(\sqrt[3]{2}z^3\right)^n\over n!}=e^{\sqrt[3]{2}z^3}$$

And the radius of convergence of the exponential is $+\infty$. If we definitely want to use the ratio test the series is the composition with $z\to z^3$ with the series

$$\sum_{n=0}^{+\infty}{2^n\over n!}Z^n\,\,\text{with}\,\,Z=z^3$$

Now apply the ratio test to the series in $Z$

$${b_{n+1}\over b_n}={2\over n+1}\to 0$$

And the radius of convergence is $+\infty$ for $Z$ and the same with $z$

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I agree the $n! z^{n!}$ term can be scary, but note that $f(z) = \sum_{n=0}^\infty n! z^{n!-1}$ is just the derivative of $g(z) = \sum_{n=0}^\infty z^{n!}$ whose radius of convergence is obvious.

For the second one, think to a well-known (entire) function.

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