Sum of powers of regularized incomplete beta function

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 $$\sum_{i=1}^{\infty}I_{p}(i,k) = \frac{k}{1-p}.$$

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