I've tried something but I'm not getting the right answer, so I'm wondering why it doesn't work.

I want to taylor expand $\frac1{z^2}$ about some point $a\in\mathbb{C}$. Here's what I did:

\begin{align*} \int\left(\frac1{z^2}\right)dz = \int\left(z^{-2}\right)dz = \frac1{-1} \frac1z = -\frac1z. \end{align*} Now expand about $z=a$, \begin{align*} -\frac1z &= -\frac1{a+z-a} = -\frac1{a\left(1+\frac{z-a}a\right)} = -\frac1a\sum_{n=0}^\infty \left(-\frac{z-a}{a}\right)^n,\\ &= \sum_{n=0}^\infty \frac{(-1)^{n+1}}{a^{n+1}}(z-a)^n.\\ \left(-\frac1z\right)' &= \frac1{z^2} = \left(\sum_{n=0}^\infty \frac{(-1)^{n+1}}{a^{n+1}}(z-a)^n\right)' = \sum_{n=0}^\infty (-1)^{n-1}\frac{n}{a^{n+1}}(z-a)^{n-1}\\ &= \sum_{n=0}^\infty (-1)^{n-1}\frac{n}{a^2}\left(\frac{z-a}{a}\right)^{n-1}. \end{align*} The binomial series, on the other hand, tells me \begin{align*} (1+x)^\alpha = \sum_{n=0}^\infty \binom{\alpha}n x^n \Rightarrow \left[1+\left(\frac{z-a}a\right)\right]^{-2} = \sum_{n=0}^\infty \binom{-2}{n} \left(\frac{z-a}a\right)^n, \end{align*} Am I doing something wrong, or is there an identity I'm not seeing?

  • $\begingroup$ Isnt the sum in $(1+x)^\alpha = \sum_{n=0}^\infty \binom{\alpha}n x^n$just go from 0 to $\alpha$? $\endgroup$ – David Szalai Nov 17 '15 at 22:51
  • $\begingroup$ @PnDChameleon Wikipedia claims it isn't: en.wikipedia.org/wiki/Binomial_series $\endgroup$ – 1010011010 Nov 17 '15 at 22:53
  • $\begingroup$ yeah, sorry, thats right for $|\alpha|<1$ $\endgroup$ – David Szalai Nov 17 '15 at 22:57
  • $\begingroup$ In your opinion, what is $\binom{-2}{n}$? $\endgroup$ – GEdgar Nov 17 '15 at 23:00
  • $\begingroup$ @GEdgar Good question.... $\endgroup$ – 1010011010 Nov 17 '15 at 23:06


By definition, we have

$$\binom{-2}{n}=\frac{(-2)(-3)(-4)\cdots (-n-1)}{(+2)(+3)(+4)\cdots \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(n)}=(-1)^n(n+1)$$

SPOLIER ALERT Scroll over the highlighted area to reveal the solution

We have \begin{align}\sum_{n=0}^\infty(-1)^{n-1}\frac{n}{a^2}\left(\frac{z-1}{a}\right)^{n-1}&=\sum_{n=0}^\infty(-1)^{n}\frac{n+1}{a^2}\left(\frac{z-1}{a}\right)^{n}\\\\&=\frac{1}{a^2}\sum_{n=0}^\infty \binom{-2}{n}\left(\frac{z-1}{a}\right)^{n}\end{align}

  • $\begingroup$ In regard to the spoiler: Why not set $m\equiv n-1$ and observe that the $m=-1$ solution doesn't contribute tot the sum? $\endgroup$ – 1010011010 Nov 17 '15 at 23:19
  • $\begingroup$ Or was that the same as what you meant in a more abstract notation...? $\endgroup$ – 1010011010 Nov 17 '15 at 23:19
  • $\begingroup$ It was the same. Note the equality of the series. On the LHS, the term at $n=0$ doesn't contribute. So, one can start at $n=1$ on the LHS. Does that make sense now? $\endgroup$ – Mark Viola Nov 17 '15 at 23:20

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