0

Can we simplify the "backward" binomial sum $$\sum_{n=k}^\infty \binom nk x^n?$$

The problem is well-known if we sum over $k$ instead of $n$, but how about the above one?

Laplacian
  • 2,494
  • 1
    It is $$\frac{x^k}{(1-x)^{k+1}},$$ for $|x|<1.$ – Thomas Andrews Dec 29 '21 at 01:29
  • 3
    $$\begin{align} \sum_{n=0}^{\infty}\sum_{k=0}^n \binom nk x^kz^n&=\sum_n (1+x)^nz^n\&=\frac{1}{1-z(1+x)}\&=\frac{1}{1-z}\frac1{1-\frac z{1-z}x}\&=\sum_{k=0}^{\infty}\frac{z^k}{ (1-z)^{k+1}}x^k\end{align} $$ – Thomas Andrews Dec 29 '21 at 01:38

1 Answers1

2

Reindex the series with $m = n-k$

$$\sum_{n=k}^\infty {n \choose k}x^n = \sum_{m=0}^\infty {m+k \choose k}x^{m+k} = x^k\sum_{m=0}^\infty {m+k \choose m}x^{m} $$

This is a negative binomial distribution with $r = k+1$

$$= \frac{x^k}{(1-x)^{k+1}}\cdot\left(\sum_{m=0}^\infty {m+k \choose m}x^{m}(1-x)^{k+1}\right) = \frac{x^k}{(1-x)^{k+1}} \cdot (1)$$

hence the name "negative binomial" since as you described it sums "backwards" from what we're used to.

Ninad Munshi
  • 34,407