# "Commutativity" of sums

In general, is $\sum_i \sum_j f(i,j) = \sum_j\sum_i f(i,j)$ ?

With $f(i,j)$ I mean some expression that depends on $i$ and $j$.

If yes, how could I prove that?

• This holds for FINITE sums. As for the proof: isn't it obvious? Addition is commutative. Jun 4, 2015 at 10:53
• @Crostul sure. What are the infinite cases? Jun 4, 2015 at 10:54
• Maybe the OP should be more specific, since its summations are dependent on some unspecified set of indices. Jun 4, 2015 at 10:55
• @SimonS Infinite cases are those that add up to infinity; these series are typically found in taylor series and such Jun 4, 2015 at 10:55
• By infinite cases I mean when the index sets for $i$ and $j$ are (countably) infinite. What's interesting then is when the sums do commute. I'm hesitating in giving an answer because I think I know a very strong sufficient condition, but I'd rather have someone chime in who works with these sorts of expressions more than I do. Jun 4, 2015 at 11:02

$$\sum_i\sum_jf(i,j) = \sum_i( f(i,j_1) + f(i,j_2)+...+f(i,j_n)) =$$$$\sum_i f(i,j_1)+\sum_if(i,j_2)+...+\sum_i f(i,j_n)=$$ $$[f(i_1,j_1)+f(i_1,j_2)+...+f(i_1,j_n)]+[f(i_2,j_1)+f(i_2,j_2)+...+f(i_2,j_n)]+...+[f(i_m,j_1)+f(i_m,j_2)+...+f(i_m,j_n)] =$$ $$[f(i_1,j_1)+f(i_2,j_1)+...+f(i_m,j_1)]+[f(i_1,j_2)+f(i_2,j_2)+...+f(i_m,j_2)]+...+[f(i_1,j_n)+f(i_2,j_n)+...+f(i_m,j_n)] =$$ $$\sum_j f(i_1,j)+\sum_j f(i_2,j)+...+\sum_j f(i_m,j) = \sum_j\sum_if(i,j).\square$$