Answer
$\sum_{j=0}^{n-2}\frac{j+1}{(n-j-1)^2}$
Work Step by Step
1. Let $j=i-1$, we have $i=j+1$ and when $i=n-1$, we have $j=n-2$
2. We can rewrite the original expression as
$\sum_{j=0}^{n-2}\frac{j+1}{(n-j-1)^2}$
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