About a sum with a combinatorial coefficient

binomial theoremcombinatoricssequences-and-seriessummation

I'm reading the Wikipedia article on analytic continuation and I got stuck in the last step of the worked example. How to get the following result? $$\sum_{n=0}^{\infty}(-1)^n{n\choose k}(a-1)^{n-k}=(-1)^ka^{-k-1}$$

Obviously the sum starts from $n=k$ because ${n\choose k}=0$ for all $k>n$. I have tried by calculating some first terms of the sum and using the identity $${n\choose k}={n-1\choose k-1}+{n-1\choose k}$$ to see if there would be a useful pattern but I have had no success. Could someone tell how to arrive to the rhs from the lhs? This question is about the same exmple and there is no answer to the op's comment. So here is the comment as a proper question.

EDIT: I had a typo on the lhs: it's actually $(-1)^n$ instead of $(-1)^k$.

Thanks to Mike Earnest's answer suggesting Newton's Binomial Theorem I calculated $$(x+1)^{-n}=\sum_{i=0}^{\infty}{-n\choose i}x^i.$$ By simplifying the binomial coefficients $$\frac{-n(-n-1)(-n-2)…(-n-i+1)}{i!}=(-1)^i\frac{n(n+1)(n+2)…(n+i-1)}{i!}=(-1)^i\frac{n+i-1}{i!(n-1)}=(-1)^i{n-1 +i\choose i}=(-1)^i{n-1+i\choose n-1}$$
and then choosing $x=(a-1)$ gives: $$\sum_{n=k}^{\infty}(-1)^n{n\choose k}(a-1)^{n-k}=a^{-k-1}.$$ So I'm only missing the $(-1)^k$. Where can that be found?

Best Answer

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

Comment:

  • In (1) we start with index $n=k$ since $\binom{n}{k}=0$ if $n<k$.

  • In (2) we shift the index to start with $n=0$ again.

  • In (3) we use the binomial identities $\binom{p}{q}=\binom{p}{p-q}$ and $\binom{-p}{q}=\binom{p+q-1}{q}(-1)^q$. We get \begin{align*} \binom{n+k}{k}=\binom{n+k}{n}=\binom{-(n+k)+n-1}{n}(-1)^n=\binom{-k-1}{n}(-1)^n \end{align*}

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

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