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A sum over one index: $\sum_i f(i)$

A sum over two indices: $\sum_i \sum_j f(i,j)$

A sum over many indices: $\sum_{k_1} \sum_{k_2} \underbrace{\dots}_n \sum_{k_n} f(\mathbf k)$?

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    $\begingroup$ I have seen the use $$\sum_{k_1,k_2,\ldots,k_n}f(\mathbf k)$$optionally with limits $0\leq k_i\leq n_i$ stacked vertically below the summation symbol. $\endgroup$ – Arthur Jan 21 '15 at 11:48
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    $\begingroup$ What @Arthur said is very common, especially in some combinatorial papers I have read. $\endgroup$ – Ali Caglayan Jan 21 '15 at 18:22
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Probably in most contexts $$\sum_{k_1} \cdots \sum_{k_n} f(k_1, \ldots, k_n)$$ suffices, and sometimes one will just write $$\sum_{k_1, \ldots, k_n} f(k_1, \ldots, k_n),$$ or perhaps slightly more precisely, $$\sum_{(k_1, \ldots, k_n)} f(k_1, \ldots, k_n).$$ As written, this last summation isn't actually a multiple summation, but rather a single summation over all $k$-tuples $(k_1, \ldots, k_n)$ in the Cartesian product $K_1 \times \cdots \times K_n$ of the index sets $K_i$ over which the index variables $k_i$ respectively vary.

As usual, one must take care when dealing with infinite sums, which formally are different objects, and for which it is possible that the summation depends on the order of indexing. In this setting, the first notation above is surely best; if you must use one of the latter notations, a comment remarking on the order of summation (or an indication that the sum is independent of the indexing order) would often be prudent.

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Try \begin{equation} \sum_{x \mathop \in S} f(x) \end{equation} which is the sum of $f(x)$ over all elements $x$ in the set $S$. Your set $S$ could be $\{k_{1},...,k_{n}\}$

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    $\begingroup$ But the OP's summations are equivalent to sum of ordered tuples, not unordered tuples, so in this case $S$ could be $\{(k_1,\ldots,k_n)\}$. $\endgroup$ – Stan Liou Jan 21 '15 at 15:46

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