Closed form of $\sum_{i=k}^\infty i h {i \choose {k-1}} h^{k-1} (1-h)^{i - (k-1)}$?

Question

Is there a closed form solution to the expression below? Or, if there is no closed form solution but the series converges, is there some upper bound on this expression?

$$\mathbb E_{i \sim Q}[i] = \sum_{i=k}^\infty i Q(i) = \sum_{i=k}^\infty i h {i \choose {k-1}} h^{k-1} (1-h)^{i - (k-1)}$$

The assumptions on constants $k$ and $h$ are that $k > 0$ and $0 < h < 1$.

I'm out of my depth here, so any tips would be appreciated!

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P.S. This is my first question on MathOverflow. Let me know if you need more context, or if there's anything I can do to improve the question itself.

Answer 1

Consider an infinite sequence of independent coin-tosses with probability $h$ of "heads" in each one. $h {i \choose k-1} h^{k-1} (1-h)^{i-(k-1)}$ is the probability that in the first $i$ coin-tosses there are exactly $k-1$ heads and the $i+1$'th toss is also heads, i.e. that the $k$'th heads occurs on the $i+1$'th toss. Your sum (if you started at $k-1$ rather than $k$) would thus be the expected number of tosses before (and not including) the $k$'th "heads". This is a standard elementary probability problem: look up "Negative binomial distribution".

Answer 2

$$ \begin{split} \sum_{i=k}^{\infty} i h \binom{i}{k-1} h^{k-1} (1-h)^{i-k+1} &= h^k (1-h)^{2-k} \sum_{i=k}^{\infty} i \binom{i}{k-1} (1-h)^{i-1} \\ &= h^k (1-h)^{2-k} f(1-h), \end{split} $$ where $f(t) = \sum_{i=k}^{\infty} i \binom{i}{k-1} t^{i-1}$. We have (as formal power series) $$ \int f \, dt = \sum_{i=k}^{\infty} \binom{i}{k-1} t^i = t^{k-1} \sum_{j=1}^{\infty} \binom{j+k-1}{k-1} t^j = \frac{t^{k-1}}{(1-t)^k} - t^{k-1} $$ (the subtraction is because the sum starts at $i=k$ instead of $k-1$). So $$ f = \frac{d}{d t} \left( \frac{t^{k-1}}{(1-t)^k} - t^{k-1} \right), $$ whatever that is, and your sum is $h^k (1-h)^{2-k} f(1-h)$.