$$ \sum_{n = 0}^\infty \frac{(4n)!}{(2n)!}k^n $$ It looks like a hypergeometric function, but a little bit different. Is there a specific name for this series or any function for this?
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2Mathematica 9.0 says the sum does not converge. – mike Mar 06 '16 at 15:49
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1The other one $\sum_{n = 0}^\infty \frac{(2n)!}{(4n)!}k^n=, _1F_2\left(1;\frac{1}{4},\frac{3}{4};\frac{k}{64}\right)$ – Claude Leibovici Mar 06 '16 at 15:53
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Apply the ratio test (assuming $\;k>0\;$ :
$$\frac{(4n+4)!k^{n+1}}{(2n+2)!}\frac{(2n)!}{(4n)!k^n}=k\frac{(4n+1)(4n+2)(4n+3)(4n+4)}{(2n+1)(2n+2)}\xrightarrow[n\to\infty]{}\infty$$
so the series can't be convergent. With the factorials inverted the above also shows it is convergent.
DonAntonio
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