"Backward" binomial sum $\sum_{n=k}^\infty \binom nk x^n$ [duplicate]

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?

Solution 1:

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.