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We assume $ a_j=f(j) $, where $ \frac{\alpha}{j}\leq f(j)\leq c $ ($ \alpha>1 $ and $0<c<1$) for $ j $ large enough. Basically, I want to calculate the order of the infinite series \begin{eqnarray*} \pi_j=\sum_{k=j}^{\infty}\frac{A_{j,k}}{(1-A_{j,k})^2}, \end{eqnarray*} where $ A_{j,k}=\Pi_{j\leq l\leq k}(1-a_l) $.

My question is how to get the general order of $ \pi_j $ (represented in terms of $ f(j) $) when $ j $ is large enough.

I have a conjecture that the order of $ \pi_j $ may be exactly $ [f(j)]^{-2} $.

The reason is that for the two extreme cases, I can calculate the order. When $ f(j)=\frac{\alpha}{j} $, the order of $ \pi_j $ is $ j^2 $. When $ f(j)=c $, the order of $ \pi_j $ is $ 1 $, which follows my conjecture. In fact, note the order of the first term is $ \frac{1-a_j}{a_j^2}=[f(j)]^{-2} $. For the two special cases, the sum of the series would have the same order as the first element. So for general case, I guess the sum would still be the same order as the first term, which is $ [f(j)]^{-2} $.

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