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Consider the following series expression:

Series[1/(a+b x^(1/3)+c x^(4/3)),{x,0,1}]

The result comes out appropriately:

1/a - b x^(1/3)/a^2 +b^2 x^(2/3)/a^3 - b^(3) x/a^4 + O[x]^(5/3)

In comparison with this, now consider the following series expansion:

Series[1/(a+b x^(Zeta[3])+c x^(1+Zeta[3])),{x,0,1}]

Here the function breaks down and gives back:

1/(a+b x^(Zeta[3])+c x^(1+Zeta[3]))

I wonder why the expansion works in the case of 1/3 but does not work for Zeta[3]? Is there a convenient way to obtain the expansion for the second case above? As always, the examples are oversimplified, so that the expansion is actually needed for expressions that are a lot bigger and unwieldy. Thanks for any suggestion!

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  • $\begingroup$ From the documentation, Details and Options, first item: "Series can construct standard Taylor series, as well as certain expansions involving negative powers, fractional powers, and logarithms." For some explanation, see comments from this thread. $\endgroup$ – Daniel Lichtblau Oct 14 '14 at 0:58
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Zeta[3] is not proper number. so I think you should apply with Rationalize[Zeta[3], 0.01] in the power position.

z3 = Rationalize[Zeta[3], 0.01];
Series[1/(a + b x^(z3) + c x^(1 + z3)), {x, 0, 3}]

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  • $\begingroup$ Wow, this works incredibly well as an approximation. Thank you! $\endgroup$ – Kagaratsch Oct 14 '14 at 17:24

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