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I need to find the interval of convergence of the following power series using either the ratio test , integral test or comparison test. Using the ratio test I found that it will converge for $ -4 < x < 4 $ but it proves inconclusive at $ x = \pm 4 $
$$ \sum_{n=0}^{\infty} \frac{(n!)^2 x^n}{(2n)!} $$

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    $\begingroup$ Do you mean $(2n)!$ on the denominator? $\endgroup$ Jun 1, 2014 at 10:01
  • $\begingroup$ oh yes . Im sorry. Ive edited it now. $\endgroup$
    – alex
    Jun 1, 2014 at 10:14

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When $x=\pm 4$, $$ |\frac{(n!)^2}{(2n)!} x^n|=\frac{n! 2^n }{1\cdot 3\cdot \cdots (2n-1)}=\frac{2\cdot 4\cdot \cdots (2n)}{1\cdot 3\cdot \cdots (2n-1)} $$ This does not converge to zero, so the series is divergent at both points.

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  • $\begingroup$ Please could you show me how you arrived at the 2nd step from the 1st? I get that you first cancelled $n!$ from the numerator and denominator but that leaves $(n+1)(n+2)(n+3)...(n+n)$ in the denominator. Then how does it give $1\cdot3...(2n-1)$ when further divided by $2^n$ ? $\endgroup$
    – alex
    Jun 2, 2014 at 8:34
  • $\begingroup$ Split the $(2n)!$ two parts: product of even numbers which is $2^n n!$, and product of odd numbers $1\cdot 3 \cdots (2n-1)$. $\endgroup$ Jun 2, 2014 at 8:42

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