3
$\begingroup$

$\frac14+\frac{x-4}{2!x^2}-\frac{(x-4)(2x-4)(3x-4)}{4!x^4}+\frac{(x-4)(2x-4)(3x-4)(4x-4)(5x-4)}{6!x^6}\mp\ldots$

Can anyone deduce the sum of this series?

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    Hehe you cheated lol. Stupid Wolfram didn't though surprisingly. But does that mean it's not original?2012-10-26

1 Answers 1

8

This is

$ -\frac1x\sum_{k=0}^\infty\binom{2k-\frac4x}{2k}\frac1{2k-\frac4x}(-1)^k\;. $

With

$ s(q,n):=\sum_{k=0}^\infty\binom{k+n}k\frac{q^k}{k+n}=\frac{(1-q)^{-n}}n\;, $

we have

$ \begin{align} -\frac1x\sum_{k=0}^\infty\binom{2k-\frac4x}{2k}\frac1{2k-\frac4x}(-1)^k &=-\frac1{2x}\left(s\left(\mathrm i,-\frac4x\right)+s\left(-\mathrm i,-\frac4x\right)\right) \\ &=-\frac1{2x}\left(\frac{(1-\mathrm i)^{4/x}}{-4/x}+\frac{(1+\mathrm i)^{4/x}}{-4/x}\right) \\ &=\frac18\left((1+\mathrm i)^{4/x}+(1-\mathrm i)^{4/x}\right) \\ &=\frac182^{2/x}\left(\mathrm e^{\mathrm i\pi/x}+\mathrm e^{-\mathrm i\pi/x}\right) \\ &=4^{1/x-1}\cos\frac\pi x\;,\end{align} $

as Alex rightly stated in a comment.

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    @joriki: The Golden Ratio.2012-10-28