You can interchange the order of summation provided all the terms are nonnegative; in this case they are (this is a special case of Tonelli's theorem).
This gives
**Your sum** $ = \sum_{j=1}^n \sum_{i=0}^{\infty} (\dfrac{j}{j+1})^i = \sum_{j=1}^n \dfrac{1}{1-\dfrac{j}{j+2}} = \frac{1}{2} \sum_{j=1}^n (j+2) = \frac{1}{2}(2n +\frac{1}{2}n(n+1)) = \frac{1}{4}n(n+5).$