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. 



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


