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I've got an expression,

$$ \left( \prod^{n}_{i = 1} \mu a_i \right) \left(\sum^{n}_{j=1} \frac{1 - a_j}{\prod^{j}_{k=1} \mu a_k} \right) $$

Where each $0 < a_j < 1 / \mu$ and $\mu > 1$, so for all $a_j$, $0 < 1 - a_j$.

I'm trying to find an analytic solution, or determine if one exists, for this expression. I know beforehand that this converges as $n \rightarrow \infty$. Both the case of finite $n$ and for infinite $n$ terms are useful.

What methods or approaches are helpful here? Pointing to similar problems is also helpful. I'm not sure what to study to understand when looking for simplifications of sums of products.

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    $\begingroup$ What do you mean by "converges"? These are just finite sums and products.... $\endgroup$ Nov 18, 2021 at 21:43

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Here's a simplification for the partial sum for you. Let $b_j = \mu a_j \in (0,1)$, and define $L_{n} := \sum_{j=1}^n \prod_{k=j+1}^n b_k$ and $M_{n} := \sum_{j=1}^n \prod_{k=j+1}^n b_{k-1}$.

Then your infinite sum is equal to $$\lim_{n\rightarrow\infty} L_{n} - \lim_{n\rightarrow\infty} \frac{b_n}{\mu}M_{n}.$$

Note that $M_n$ is just $L_n$ with $b_i \rightarrow b_{i-1}$ so you only need to calculate one of the two terms. This looks to me to be quite a bit simpler than your original form for the sum.

If you define $x := \min_{k=2,...,n}\{b_k\}$ and $y = \max_{k=2,...,n} \{b_k\}$ then using the geometric series I believe you have $$\frac{1-x^n}{1-x} \leq L_n \leq \frac{1-y^n}{1-y},$$ which may be relevant in understanding convergence of the limit.

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    $\begingroup$ Thank you for your insight. Your rearrangement is the trick I needed. The coefficients of $a_j$ are a stand in for a functional form, so expressing this sum as the limit of $L_n$ is something I can do. Thank you! $\endgroup$
    – abnowack
    Nov 19, 2021 at 0:18
  • $\begingroup$ Glad to help. I just realised I'd got a + instead of - typo, I edited to fix it. Maybe you already picked up on this but if you didn't then make sure to have another look at my answer. $\endgroup$
    – Jojo
    Nov 19, 2021 at 8:35

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