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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    shouldn't the first term be 4, so one can have a uniform pattern?2012-10-26
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    @F'OlaYinka: I think it's OK as it is.2012-10-26
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    It's equal to $4^{\frac{1}{x}-1} \cos\big[\frac{\pi}{x}\big]$2012-10-26
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    is the series well defined this way? $\displaystyle \cfrac 14 + \sum_{k\ge1} (-1)^{k+1}\cfrac { \prod_{n=1}^{2k-1} (nx -4)} {(2k)! \space x^{2k}}$2012-10-26
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    @F'OlaYinka $\prod_{n=1}^{2k-1}(nx-4)$ :)2012-10-26
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    @J.G. Thanks! fixed it.2012-10-26
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    Ok Alex is right. How did you know Alex?2012-10-26
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    My dear friend Mathematica knew it :)2012-10-26
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    Maple doesn't seem to :(2012-10-26
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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