About a sum with a combinatorial coefficient

Question

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?

Answer 1

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.

Answer 2

You need to use this general fact: $$ \sum_{n= 0}^\infty \binom nkx^n = \frac{x^k}{(1-x)^{k+1}}\qquad (|x|<1) $$ In your case, you want $x=a-1$.

To prove this, let $S_k=\sum_{n\ge 0}\binom nkx^k$. Using Pascal's identity, $\binom nk=\binom{n-1}{k-1}+\binom {n-1}k$, you can split this into two sums, and derive the equation $$ S_k = xS_{k-1}+xS_k, $$ which, combined with the base case $S_0=(1-x)^{-1}$, quickly leads to a proof by induction that $S_k=x^k(1-x)^{-(k+1)}$.

This also follows from Newton's binomial theorem: $$ \sum_{n\ge 0}\binom n{n-k}x^n =\sum_{n\ge 0}(-1)^{n-k}\binom{-(k+1)}{n-k}x^n =x^k\sum_{n\ge 0}\binom{-(k+1)}{n-k}(-x)^{n-k}=x^k(1-x)^{-(k+1)} $$