# sum of infinite series with each element containing infinite product

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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