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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!


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.

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    $\begingroup$ It's weird that $i$ appears as the index in the summation, and appears in the left-hand term as well (which sounds cryptic to me). $\endgroup$ – YCor Jun 7 '17 at 22:18
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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".

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$$ \begin{split} \sum_{i=k}^{\infty} i h \binom{i}{k-1} h^{k-1} (1-h)^{i-k+1} &= h^k (1-h)^{2-k} \sum_{i=k}^{\infty} i \binom{i}{k-1} (1-h)^{i-1} \\ &= h^k (1-h)^{2-k} f(1-h), \end{split} $$ 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 $h^k (1-h)^{2-k} f(1-h)$.

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