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?

  • 1
    $\begingroup$ This holds for FINITE sums. As for the proof: isn't it obvious? Addition is commutative. $\endgroup$
    – Crostul
    Jun 4, 2015 at 10:53
  • $\begingroup$ @Crostul sure. What are the infinite cases? $\endgroup$
    – Simon S
    Jun 4, 2015 at 10:54
  • $\begingroup$ Maybe the OP should be more specific, since its summations are dependent on some unspecified set of indices. $\endgroup$
    – Crostul
    Jun 4, 2015 at 10:55
  • $\begingroup$ @SimonS Infinite cases are those that add up to infinity; these series are typically found in taylor series and such $\endgroup$ Jun 4, 2015 at 10:55
  • $\begingroup$ 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. $\endgroup$
    – Simon S
    Jun 4, 2015 at 11:02

1 Answer 1


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

  • 2
    $\begingroup$ As discussed in the comments above, this finite case is trivial. $\endgroup$
    – Simon S
    Jun 4, 2015 at 11:16
  • $\begingroup$ I guess there is a misprint in the derivation above in the line "3" where you expanded in "j" not in "i". $\endgroup$ Sep 16, 2021 at 16:25

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