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Let $k$ and $r$ be natural numbers, and $p$ is a rational number in $[0,1]$. Is it possible to compute exactly in closed form the following sum?

$$ \sum_{i=1}^{\infty}I_{p}(i,k)^r, $$ where $I_x(a,b)$ is the regularized incomplete beta function.

[Update 6 Feb 2019]

Motivation for $r=1$

Let $X$ be a random variable distributed according to the negative binomial distribution with parameters $(i-k; k, 1-p)$, where the probability mass function of the negative binomial distribution with parameters $(k; r, p)$ is $f(k;r,p) = \Pr(X=k) = {r+k-1 \choose k} p^k (1-p)^r$. It is known that $$ \Pr(X \leq i) = 1- I_{1-p}(i+1,k). $$

On the other hand, we know that the variable $X$ is the sum of $k$ independent geometrically distributed variables with parameter $p$. Hence

$$ \mathbb{E}[X] = \sum_{i=k}^{\infty} i \Pr(X_{\mathcal{P}} = i) = \sum_{i=0}^{\infty} \Pr(X_{\mathcal{P}} > i) = \sum_{i=0}^{\infty} (1-\Pr(X_{\mathcal{P}} \leq i)) = \sum_{i=0}^{\infty} I_{1-p}(i+1,k) = \frac{k}{p}. $$

Therefore for $r=1$ we have $$ \sum_{i=1}^{\infty}I_{p}(i,k) = \frac{k}{1-p}. $$

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    $\begingroup$ I doubt it even having a closed form for the simple case of $r=1$. $\endgroup$ – Ali Feb 1 at 18:12
  • $\begingroup$ @Ali, interestingly, for $r=1$ it can be computed in closed form. See my update. $\endgroup$ – Victor Feb 6 at 13:39

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