question about double sums

Question

Suppose we have an expression

$$ \sum_{1 \leq k < j \leq n } f(k)f(j) $$

Can we express this as a double sum like

$$ \sum_{k=1}^n \sum_{j=1}^n f(k)f(j) $$

???

Answer 1

I'd rather say: $$\sum_{1\leq k\leq j\leq n}f(j)\,f(k)=\frac{1}{2}\left(\left(\sum_{h=1}^{n}f(h)\right)^2+\sum_{h=1}^{n}f(h)^2\right),$$ $$\sum_{1\leq k< j\leq n}f(j)\,f(k)=\frac{1}{2}\left(\left(\sum_{h=1}^{n}f(h)\right)^2-\sum_{h=1}^{n}f(h)^2\right).$$

Answer 2

Not quite. As $k$ ranges from $1$ to $n-1$, $j$ ranges from $k+1$ to $n$. So you should have $$\sum_{k = 1}^{n-1} \sum_{j = k+1}^n f(k)f(j)$$

Answer 3

Say $n=3$. The original sum contains only the red terms in this matrix: $$ \begin{bmatrix} f(1)f(1) & \color{red}{f(1)f(2)} & \color{red}{f(1)f(3)} \\ f(2)f(1) & f(2)f(2) & \color{red}{f(2)f(3)} \\ f(3)f(1) & f(3)f(2) & f(3)f(3) \\ \end{bmatrix} $$ However, your proposed sum contains all of the terms in the matrix!

This isn't hard to fix, though. Notice that the matrix is symmetric because $f(j)f(k) = f(k)f(j)$, so the original sum is half your sum, once you get rid of the diagonal terms: $$ \sum_{1 \leq k < j \leq n } f(k)f(j) = \frac{1}{2}\left[ \sum_{k=1}^n \sum_{j=1}^n f(k)f(j) - \sum_{k=1}^n f(k)f(k)\right]. $$

Then you get Jack's answer by recognizing that $$ \sum_{k=1}^n \sum_{j=1}^n f(k)f(j) = \sum_{k=1}^n f(k) \times \sum_{j=1}^n f(j)=\left( \sum_{k=1}^n f(k) \right)^2. $$

Answer 4

No. You haven't taken into account the middle inequality in $1\leq k<j\leq n$. So your second sum should be $$\sum_{k=1}^n\sum_{j=k+1}^nf(k)f(j).$$