Sylvester's sequence is defined recursively with $e_1 = 2$ and $e_{n + 1} = e_n^2 - e_n + 1$. The sequence also satisfies the recurrence $e_{n + 1} = e_1 \cdots e_n + 1$. These recurrences let us compute the remarkable sum $$\sum_{k = 1}^n \frac{1}{e_k} = \frac{e_{n + 1} - 2}{e_{n + 1} - 1}.$$ For instance, compute $R_{n + 1} - R_n$ where $R_n = (e_n - 2) / (e_n - 1)$, then sum over $n$.
Question: In general, given a sequence $\{a_n\}$ satisfying the equivalent recurrences \begin{align*} a_{n + 1} &= \prod_{k = 1}^n a_k + \beta \\ a_{n + 1} &= a_n^2 - \beta a_n + \beta, \end{align*} can we evaluate $\sum_{k = 1}^n a_k^{-1}$ in closed-form? This is similar to this question, but more restricted.
I suspect that the answer is nice because of the following calculation: \begin{align*} \sum_{k = 1}^n a_k^{-1} &= \sum_{k = 1}^n \frac{a_1 \cdots a_{k - 1} a_{k + 1} \cdots a_n}{a_1 \cdots a_n} \\ &= \frac{P_n}{a_{n + 1} - \beta}. \end{align*} The sequence $P_n$ is some messy sum that I believe could reduce.
Using a trick similar to the one above, it is "routine" to show that $$\sum_{k = 1}^n \frac{(\gamma - \beta)(a_k - 1)}{(a_k - \beta) a_k} = \frac{a_{n + 1} - \gamma}{a_{n + 1} - \beta} - \frac{a_1 - \gamma}{a_1 - \beta}$$ for any $\gamma$. When $\beta = 1$ we can take $\gamma = 2$ to evaluate $\sum_{k = 1}^n a_k^{-1}$, but this only works for $\beta = 1$.