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Is there an alternate way to represent the multiple summation given below?
$\displaystyle \Large \sum_{i_k=k}^{n} \space \sum_{i_{k-1}=k-1}^{i_k} \dots \sum_{i_2=2}^{i_3} \space \sum_{i_1=1}^{i_2}$
It guess it is wrong to write it as a product of summations like $\displaystyle \large \prod_{\Large j=k}^{1}\space \sum_{\Large i_j=j}^{\Large i_{j+1}}$.

What's a better, concise notation ?

Update: Another way is $\displaystyle \quad \large \sum\limits_{\Large n \geq i_k \geq i_{k-1} \dots \geq i_2 \geq i_1 \geq 1}$.

Are there any other ways?

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  • $\begingroup$ Note that the update isn't correct, since it allows for $(1,1,1,\ldots)$ which your original notation doesn't allow for. $\endgroup$ – Calvin Lin Jun 12 '13 at 14:26
  • $\begingroup$ yea I missed one constraint. The answer by Goos seems to be fine. $\endgroup$ – AIB Jun 12 '13 at 15:09
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I would use $$ \large \sum_{\substack{1 \le i_1 \le i_2 \le \cdots \le i_k \le n \\ i_j \ge j \;\; \forall j}} $$

A summation in this form is understood to be taken over all ordered tuples $(i_1,i_2,\cdots,i_k)$ which satisfy the inequalities given. See Wikipedia for a list of various notational conventions.

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  • $\begingroup$ @CalvinLin $i_j \ge j$ is in my condition. Perhaps I misunderstand you (the OP copied down my sum incorrectly). $\endgroup$ – 6005 Jun 12 '13 at 14:30
  • $\begingroup$ Ah, my bad, I didn't see that you had it. I just saw OP's version. $\endgroup$ – Calvin Lin Jun 12 '13 at 14:31
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    $\begingroup$ $\displaystyle\sum_{l \le i_l \le i_{l+1}}_{ 1 \le l \le k}$... Is this type of notation accepatable? with $i_{k+1}=n$? $\endgroup$ – AIB Jun 25 '13 at 22:17

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