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I want to find the radius of convergence for the series in z where the coefficient of $z^n$ is 1). $a_n =\frac{(n!)^3}{(3n)!}$

Also I need to find the radius of convergence of

2). hypergeometric series.

$F(\alpha, \beta, \gamma;z)=1+ \sum \frac {\alpha (\alpha +1)...(\alpha + n -1) \beta (\beta +1)...(\beta n -1)}{n! \gamma (\gamma + 1) ...(\gamma + n -1)}. z^n$

Let f(z)= $\sum a_n z^n$ I know that for a holomorphic function $f$ whose power series has coefficient $a_n$ is gi ven as

$$\frac{1}{R}= \lim_{x \to \infty} \left|\frac{a_{n+1}}{a_n}\right|$$

For 1) I got the answer R=27. Am I right?? I have no idea for hypergeometric series.

  • 2
    For 1), 27 is right answer. But to get good guidance here, it is always recommended to show your work. Also in case of 2) You should write down the expression for a geometric series. Use [this](http://meta.math.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference) link to format your post. Cheers! – 2017-02-21
  • 0
    Yes I wrote down the expression for hypergeometric series. Need hint. – 2017-02-21
  • 1
    $\frac 1R=\lim_{n \to \infty} |\frac {(\alpha + n)(\beta + n)}{(n+1)(\gamma +n)}|$ – 2017-02-21

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