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I know that

$$\displaystyle \sqrt{1+x} = \sum_{j=0}^{\infty}\left( \frac{(-1)^{(j-1)}}{2^{2j-1}\cdot(2j-1)}\binom{2j-1}{j}x^j\right). $$

Now, I want to evaluate $\sqrt[3]{1+x}$ but stuck at some point:

To evaluate $\sqrt[3]{1+x}$, I first tried to find what $\binom{1/3}{j}$ is

$$\binom{1/3}{j} = \frac{(1/3 \cdot (1/3-1) \cdot (1/3-2) \cdots (1/3-j+1))}{j!} = \frac{(-1)^{j-1}\cdot 1\cdot 2\cdot 5\cdots (3j-4)}{3^j\cdot j!}$$

However I could not continue from here. Is there a way to write $1\cdot 2\cdot 5\cdots(3j-4)$ in another form? (Or is $3\cdot 4\cdot 6\cdot 7\cdots(3j-5)$ possible?) Could you please help me?

Regards

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    $\begingroup$ I slightly improved the $\LaTeX$ in your question. Please check that I kept the meaning of the question. $\endgroup$ Commented Mar 18, 2013 at 23:02
  • $\begingroup$ @GitGud thank you so much! $\endgroup$ Commented Mar 18, 2013 at 23:09
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    $\begingroup$ I don't think you can write anything for $(2)(5)\cdots(3j-4)$. Why do you feel you need to? $\endgroup$ Commented Mar 18, 2013 at 23:23
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    $\begingroup$ If there isn't a solution, there isn't a problem. Not every complex mathematical expression has a simpler form. I think you have to live with what you have. $\endgroup$ Commented Mar 18, 2013 at 23:31
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    $\begingroup$ @Xentius it's called gamma function. $\endgroup$ Commented Mar 19, 2013 at 7:53

1 Answer 1

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Note that $\binom{n}{r} = \frac{n!}{r!(n-r)!}$. We also have the identity $\Gamma(n) = (n-1)!$

Thus, we can generalize the binomial coefficient as:

$$\binom{n}{r} = \frac{\Gamma(n+1)}{\Gamma(r + 1)\Gamma(n-r + 1)}$$

Therefore: $$\binom{1/3}{j} = \frac{\Gamma(4/3)}{\Gamma(j+1)\Gamma(j + 2/3)}$$

Note, though, that you may run into some issues, as $r$ (as in, $\binom{n}{r}$) should be in the range $[0, n]$.

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