Can this double sum be simplified?

Question

Can this summation be simplified ($A$ is a constant)?

$$\displaystyle\sum_{i=1}^{A} \displaystyle\sum_{j>i}^{A} f(i)f(j)$$

By simplified I mean either a closed-form expression in terms of $A$ and $f$, or failing that even just removing one of the sums.

If not, what if the $j>i$ is replaced by $j=1$?

Answer 1

• $$\sum_{i=1}^{A} \sum_{j>i}^{A} f(i)f(j) =\sum_{i=1}^{A} f(i)\sum_{j=i+1}^{A} f(j) $$but you can also swap the indices: $$ \sum_{i=1}^{A} \sum_{j>i}^{A} f(i)f(j) = \sum_{j=1}^{A} f(j)\sum_{i=j+1}^{A} f(i) = \sum_{i=1}^{A} f(i)\sum_{j=1}^{i-1} f(j) $$ Hence, $$ 2 \sum_{i=1}^{A} \sum_{j>i}^{A} f(i)f(j) = \sum_{i=1}^{A} f(i)\sum_{j\neq i} f(j) =\sum_{i=1}^{A} f(i)\left[\left(\sum_{j=1}^A f(j)\right) -f(i) \right] \\= \left(\sum_{j=1}^A f(j)\right)^2 - \sum_{i=1}^{A} f(i)^2. $$
• $$\sum_{i=1}^{A} \sum_{j=1}^{A} f(i)f(j) = \sum_{i=1}^{A} f(i)\sum_{j=1}^{A} f(j) = \left(\sum_{j=1}^A f(j)\right)^2$$