closed form for $\sum_{k=0}^{v-1}\binom{n-v-1}{k}(\frac{p}{q})^k$

Question

In an exercise involving probabilty, in which $p=1-q$, and $v$ is a given positive integer I try to show that

$\displaystyle \sum_{n=0}^{+\infty} t^v p^v (1-p)^{n-v} \sum_{k=0}^{v-1}\binom{n-v-1}{k}\left(\frac{p}{q}\right)^k=\frac{p^vt^v(1-pt)}{1-t+p^vqt^{v+1}}$.

I think that if I'm able to give a simplified expression of $\displaystyle\sum_{k=0}^{v-1}\binom{n-v-1}{k}\left(\frac{p}{q}\right)^k$ it could help. Any idea ?

This series gives the law of the variable giving the number of independent Bernoulli experiments needed to obtain $v$ successive favorable outcomes.

Answer 1

If $p/q=x$, Mathematica gives: $$(1 + x)^{-1-v} ((1 + x)^ n - (1 + x) (x (1 + x))^ v{-1 + n - v \choose v} Hyper_2F_1[1 - n + 2 v, 1, 1 + v, -x])$$

Answer 2

I have very strong gut feeling that the summation you are looking at does not have a closed form. This is akin to the fact that the partial sums $\sum_{k=0}^r\binom{n}k$ do not behave nicely, even though the whole sum $\sum_{k=0}^n\binom{n}k=2^n$ is nice.

However, if you reverse the order of summation, you arrive at $$ t^v\left(\frac{p}{1-p}\right)^{v}\cdot \sum_{k=0}^{v-1} \left(\frac{p}q\right)^{k}\sum_{n=0}^{\infty}\binom{n-v-1}{k}q^n $$ Here, the inner summation does have closed form. To find it, manipulate the summation until you can use the below Lemma.

Lemma:$$ \sum_{m=0}^\infty \binom{m}jx^m=\frac{x^j}{(1-x)^{j+1}} $$

Proof: \begin{align} \sum_{m=0}^\infty \binom{m}jx^m &=x^j\sum_{m=0}^\infty \binom{m}{m-j}x^{m-j} \\&=x^j\sum_{m=0}^\infty \frac{m(m-1)\cdots (j+1)}{(m-j)!}x^{m-j} \\&\stackrel1=x^j\sum_{m=0}^\infty \frac{(-j-1)(-j-2)\cdots(-m)}{(m-j)!}(-x)^{m-j} \\&\stackrel2=x^j\sum_{m=0}^\infty \binom{-j-1}{m-j}(-x)^{m-j} \\&\stackrel3=x^j\sum_{m=j}^\infty \binom{-j-1}{m-j}(-x)^{m-j} \\&\stackrel4=x^j\sum_{m=0}^\infty \binom{-j-1}{m}(-x)^{m} \\&\stackrel5=x^j\cdot (1-x)^{-j-1} \end{align} Explanations:

1. I reversed the order of hte factors in the numerator, and introduced a negative sign into each of them, replacing $x^{m-j}$ with $(-x)^{m-j}$ to compensate for the extra negatives.

2. The definition of $\binom{n}k$ can be extended to allow $n$ to be negative, via $\binom{n}{k}=\frac{n(n-1)\cdots(n-k+1)}{k!}$. Here, we use that generalized definition to write the fraction as a binomial coefficient.

3. The first $j$ terms are zero, so can be ignored.

4. Reindex the sum so that $m$ starts at zero.

5. This is Newton's binomial theorem, which says that $(1+x)^n=\sum_{k=0}^\infty \binom{n}kx^k$ is valid even when $n$ is negative (or any real number), as long as you use the generalized definition of $\binom{n}k$ discussed in point $2$.