# Can I integrate then differentiate this power series to derive the same result as the binomial series expansion?

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?

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

HINT:

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}

• 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? – 1010011010 Nov 17 '15 at 23:19
• Or was that the same as what you meant in a more abstract notation...? – 1010011010 Nov 17 '15 at 23:19
• 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? – Mark Viola Nov 17 '15 at 23:20