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Clearly it is permitted to put restrictions on the indices in a summation, and

One often sees generalizations of this notation in which an arbitrary logical condition is supplied, and the sum is intended to be taken over all values satisfying the condition. Here are some common examples: $$\sum_{0\le k< 100} f(k)$$ is the sum of $f(k)$ over all (integers) $k$ in the specified range, $$\sum_{x \mathop \in S} f(x)$$ is the sum of $f(x)$ over all elements $x$ in the set $S$

However, if I don't have a symbol for the elements of my collection $E$, but rather address them as $E_i$, is it acceptable notation to index into the collection both in the condition and in the summand? $$1\over\displaystyle\sum_{\substack{i=1 \\ E_i\neq 0}}^n\frac1{E_i}$$ If this is acceptable, it should give the same sum as this verbose use of Iverson brackets: $$1\over\displaystyle\sum_{i=1}^n\frac{[E_i≠0]}{E_i+[E_i=0]}$$

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  • $\begingroup$ It's fine. Don't use the Iverson brackets. $\endgroup$ – Ethan Bolker Nov 7 '19 at 22:04
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You don't actually need the condition in the summation, nor the verbose use of Iverson brackets: $$1\over\displaystyle\sum_{i=1}^n\frac1{E_i}[E_i\neq 0]$$ This succinctly expresses the idea of summing the reciprocals of the non-zero elements. Now, you might counter that this involves the undefined $1/0$ when $E_i=0$, but, in the context of the sum $\sum_k a_k\bigl[P(k)\bigr]$, Concrete Mathematics says that

Sometimes $a_k$ isn't defined for all integers $k$. We get around this difficulty by assuming that $\bigl[P(k)\bigr]$ is “very strongly zero” when $P(k)$ is false; it’s so much zero, it makes $a_k\bigl[P(k)\bigr]$ equal to zero even when $a_k$ is undefined.

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  • $\begingroup$ We usually write $E_i \ne 0$ in the index part for better readability. The index part tells us something about the players right at the beginning and we can exclude this way $E_i$ which would otherwise be involved in (maybe complicated) expressions. We use Iverson brackets as part of summands when we use them actually for calculation. Some examples are provided in the referred book. $\endgroup$ – Markus Scheuer Nov 8 '19 at 8:23
  • $\begingroup$ @MarkusScheuer Since there's no obviously correct way to specify both the exact indices and a condition on the value at the current index, then this seems to be an acceptable solution since an authority uses it, though it may not be ideal. Feel free to revive your answer :-) $\endgroup$ – Adám Nov 8 '19 at 11:11
  • $\begingroup$ @Adáms: Thanks. To me it seems, you could already do a fruitful step in the direction you wanted and you also incorporated my reference to one of the great. So, a revival of my answer isn't needed. Best, $\endgroup$ – Markus Scheuer Nov 8 '19 at 11:19
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I don't think it's a standard notation, but on the other hand I can hardly see how anybody could misinterpret that notation. One could use a more standardized notation, but that might suffer from it not being that easy to read, for example writing the entire condition below the sigma (ie $i\in\mathbb N\land 1\le i\le n\land E_i\ne 0$).

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