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How to find the radius of convergence and interval of convergence of the following power series?

$\sum{x^{n!}}$

and

$ \sum{\frac{1}{n^{\sqrt{n}}} x^n} $

  • 0
    A [related question](http://math.stackexchange.com/q/228852/26872).2012-12-08

2 Answers 2

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For the first one just use the ratio test:

$\frac{x^{(n+1)!}}{x^{n!}}=x^{(n+1)!-n!}=x^{n!((n+1)-1)}=x^{n\cdot n!}\;.$

Now what’s $\lim_{n\to\infty}|x|^{n\cdot n!}\;?$

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    @Babak: Shh! ;-) Yes, it is.2012-12-08
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Hint: Use the root test for the second series and note that

$ \lim_{n\to \infty} \left(\frac{1}{n^{\sqrt{n}}}\right)^{1/n}=1. $

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    sorry,sir.I apologize..2012-12-08