Simplifying an expression using the binomial theorem

Question

I am going through the analytic continuation article here. I am working through the worked example but am stuck at the final step.

Here is the problem:

$$a_{k} =\frac{1}{2\pi}\sum_{n = 0}^{\infty} (-1)^{n}\int_{0}^{2\pi} \sum_{m=0}^{n}\binom{n}{m}(a-1)^{n-m}(re^{i\theta})^{m-k}d\theta.$$

But since $\int_{0}^{2\pi}(re^{i\theta})^{m-k}d\theta = 2\pi r^{m-k} \delta_{m,k}$, only the $m = k$ term in the finite summation survives and we are left with:

$$a_{k} = \sum_{n = 0}^{\infty} \binom{n}{k}(-1)^{n}(a-1)^{n-k}$$

How do I get from here to

$$a_{k} = (-1)^{k}a^{-k-1}? $$

Answer 1

We obtain \begin{align*} \color{blue}{\sum_{n=0}^\infty}&\color{blue}{\binom{n}{k}(-1)^n(a-1)^{n-k}}\\ &=\sum_{n=k}^\infty\binom{n}{k}(-1)^n(a-1)^{n-k}\tag{1}\\ &=\sum_{n=0}^\infty\binom{n+k}{n}(-1)^{n+k}(a-1)^{n}\tag{2}\\ &=(-1)^k\sum_{n=0}^\infty\binom{-k-1}{n}(a-1)^{n}\tag{3}\\ &\,\,\color{blue}{=(-1)^ka^{-k-1}}\tag{4}\\ \end{align*} and the claim follows.

Comment:

• In (1) we set the lower limit to $n=k$, since $\binom{n}{k}=0$ if $n<k$.

• In (2) we shift the index to start with $n=0$. We also use $\binom{p}{q}=\binom{p}{p-q}$.

• In (3) we use the binomial identity $\binom{-p}{q}=\binom{p+q-1}{q}(-1)^q$.

• In (4) we apply the binomial series expansion.