# Does $\sum_{i\neq j}$=$\sum_{i=1}^{n}\sum_{j=1}^{n}$?

I want to know if $\sum_{i\neq j}$=$\sum_{i=1}^{n}\sum_{j=1}^{n}$?

• What does the first sum (left-side) mean? Is it a single sum over $i$, or does it represent a double sum (like the right-side)? – Arturo don Juan Dec 12 '15 at 9:02
• @ArturodonJuan This would have to be the sum over $\{(i,j): 1 \le i,j \le n, i \neq j \}$ going from context. – Henno Brandsma Dec 12 '15 at 9:15

The answer is: no. We have, for example $$\sum_{1\leq i\neq j \leq 2}a_{i,j}=a_{1,2}+a_{2,1}$$ and $$\sum_{i=1}^{2}\sum_{j=1}^{2}a_{i,j}=a_{1,1}+a_{1,2}+a_{2,1}+a_{2,2}$$ which are different in general.
• What about $\sum_{i\neq j}^{n} x_i$? Is it equal to $n(x_1+...+x_n)$? – Spencer Ireland Dec 12 '15 at 9:13
No. In the twofold sum the indices $i$ and $j$ are independent, so they can asume all combinations. E..g for $n=2$ we have $$\sum_{i=1}^2 \sum_{j=1}^2 a_{i,j} = \sum_{i=1}^2 (a_{i,1} + a_{i,2}) = a_{1,1} + a_{1,2} + a_{2,1} + a_{2,2}$$
while $$\sum_{1 \le i \neq j \le 2} a_{i,j} = a_{1,2} + a_{2,1}$$
• What about $\sum_{i\neq j}^{n} x_i$? Is it equal to $n(x_1+...+x_n)$? – Spencer Ireland Dec 12 '15 at 9:08
• @SpencerIreland There is no $j$ in the term? That's strange – Henno Brandsma Dec 12 '15 at 9:13
• If we indeed sum over the set in my comment: it's $(n-1)(x_1 + \ldots +x_n)$, because every $i$ occurs $n-1$ times in the set. – Henno Brandsma Dec 12 '15 at 9:18