TITLE: Closed form of $\sum_{i=k}^\infty i h {i \choose {k-1}} h^{k-1} (1-h)^{i - (k-1)}$?
QUESTION [0 upvotes]: 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.

REPLY [0 votes]: $$
\begin{split}
  \sum_{i=k}^{\infty} i h \binom{i}{k-1} h^{k-1} (1-h)^{i-k+1}
    &= h^k (1-h)^{2-k} \sum_{i=k}^{\infty} i \binom{i}{k-1} (1-h)^{i-1} \\
    &= h^k (1-h)^{2-k} f(1-h),
\end{split}
$$
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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