Indexing twice for conditional summation 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:
$$1\over\displaystyle\sum_{i=1}^n\frac{[E_i≠0]}{E_i+[E_i=0]}$$

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. 

