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In this thread, General formula for the power series of $\dfrac{1}{(1+x)^3}$

the_candyman posted an identity:

$$(1-z)^{-\alpha} = \sum_{n=0}^{+\infty}{{n + \alpha - 1}\choose{n}}z^n.$$

Notice that this binomial series has a negative exponent. I know that the general formula for binomial series is:

$$(1+x)^k=\sum_{n=0}^{+\infty}{\binom{k}{n}}x^n$$.

How do apply the second definition to prove the first identity?

James Warthington
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    Try using the binomial series and the negative binomial identity. – Minus One-Twelfth Nov 16 '19 at 00:41
  • Put $x=z, k=-\alpha$ and expand the binomial coefficients in full and you will see the identity $$(-1)^n\binom{-\alpha}{n}=\binom{n+\alpha - 1}{n}$$ in a very obvious manner. – Paramanand Singh Nov 16 '19 at 02:07
  • @Paramanand Singh $\binom{n+k-1}{n}=\dfrac{(n+k-1)!}{k!(n+k-1-k)}=\dfrac{(n+k-1)!}{k!(n+1)!}$, is this correct, how can I simplify this further to give the right hand side? – James Warthington Nov 16 '19 at 02:14
  • You need to expand in full like $$(-1)^n\binom{-\alpha}{n}=(-1)^n\frac{(-\alpha)(-\alpha-1)\dots(-\alpha-n+1)}{n!}$$ and then you get desired result. – Paramanand Singh Nov 16 '19 at 02:17
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    @ParamanandSingh: Can you do this in full steps so I can learn, you can answer my question and I will mark your answer. This way you can earn a few points. Sorry, I haven't mastered the binomial notation yet so I am kind of slow. – James Warthington Nov 16 '19 at 02:18
  • There are no further steps. Just note that $(-1)^n$ is used to change signs of all the $n$ factors in denominator to get $$\frac{\alpha(\alpha+1)\dots(\alpha+n-1)}{n!}$$ and this is $$\binom{n+\alpha-1}{n}$$ – Paramanand Singh Nov 16 '19 at 02:21
  • The definition of binomial coefficient is $$\binom{a} {n} =\frac{a(a-1)(a-2)\dots(a-(n-1))}{n!}$$ where $n$ is a positive integer. – Paramanand Singh Nov 16 '19 at 02:23

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We use the following definition of binomial identities valid for complex $\alpha$ and non-negative integers $n$: \begin{align*} \binom{\alpha}{n}=\frac{\alpha(\alpha-1)\cdots(\alpha-n+1)}{n!}\tag{1} \end{align*} which can be found for instance as formula (5.1) in Concrete Mathematics by R.L. Graham, D.E. Knuth and O. Patashnik.

We obtain \begin{align*} \color{blue}{\binom{n+\alpha-1}{n}}&=\frac{(n+\alpha-1)((n+\alpha-1)-1)\cdots(n+\alpha-1-(n-1))}{n!}\tag{2}\\ &=\frac{(n+\alpha-1)((n+\alpha-2)\cdots\alpha}{n!}\\ &=(-1)^n\frac{(-\alpha)(-\alpha-1)\cdots(-\alpha-(n-1))}{n!}\tag{3}\\ &\,\,\color{blue}{=(-1)^n\binom{-\alpha}{n}}\tag{4} \end{align*}

Comment:

  • In (2) we use the definition (1).

  • In (3) we factor out $(-1)^n$.

  • In (4) we use the definition (1) again.

From (2) and (4) we conclude \begin{align*} (1-z)^{-\alpha}=\sum_{n=0}^{\infty}\binom{-\alpha}{n}(-z)^n=\sum_{n=0}^\infty\binom{n+\alpha-1}{n}z^n\qquad\qquad |z|<1 \end{align*}

Markus Scheuer
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