The expectation of a Bernoulli divided by an associated sum of bernoullis

Question

For $i=1,\dots, k$, let $X_i \sim \mathrm{Bernoulli}(p_i)$ be independent such that $\sum_{i=1}^k p_i > 0$. How do I evaluate the following expectation: $$\mathbb{E} \left( \frac{X_i}{\sum_{j=1}^k X_j} \right),$$ where consider that $0/0=0$ by convention? I see that the denominator has a Poisson Binomial distribution. When $p_1 = \dots = p_k$, numerical evidence suggests that the expectation is $1/k$; indeed this can easily be proven by symmetry. In general, numerical evidence suggests that the expectation does not evaluate to $p_i / \sum_{j=1}^k p_j$, although I'm not sure.