Clarification of summation upper bound in $\sum_{p\text{ prime}} \frac{1}{2^p}$ I tried searching for this, but didn't know how to apply what I found to the specific problem I have. If I have the summation $$\sum_{p\text{ prime}} \frac{1}{2^p},$$ what is the implied upper bound? I think it's $\infty$ (i.e., infinite amount of primes), but I'm not very certain. 
Thanks in advance. 
 A: It is often desirable to sum over some set other than a sequence of consecutive integers.  In such a case, it is common to specify the domain of summation as a subscript to the large sigma notation.  For example, if we wanted to sum all of the even terms of a sequence $\{a_n\}$, then we might write
$$ \sum_{\{n=2m : m\in\mathbb{N}\}} a_n
\qquad\text{rather than the usual}\qquad
\sum_{m=1}^{\infty} a_{2m}. $$
More generally, given some sequence $\{a_n\}$ and a subset $E$ of the natural numbers, we can indicate our desire to add up only the terms which correspond to elements of $E$ by writing
$$ \sum_{n=1}^{\infty} a_n \chi_E(n), $$
where $$ \chi_E(n) = \begin{cases}
1 & \text{if $n\in E$, and} \\
0 & \text{otherwise}
\end{cases} $$
denotes the characteristic function of $E$.  However, this notation is quite clunky, and adds complication without adding clarity.  Using the subscript notation is much simpler:
$$ \sum_{n=1}^{\infty} a_n \chi_E(n)
= \sum_{n\in E} a_n. $$
This is the meaning of the sum given in the question, though even more notation has been stripped out.  If I wanted to be really pedantic, I might write

Let $\mathscr{P} \subseteq \mathbb{N}$ denote the set of all prime numbers.  Then
  $$ \sum_{p\in\mathscr{P}} \frac{1}{2^p} = \dotsb .$$

or, more simply, we just write
$$ \sum_{\text{$p$ prime}} \frac{1}{2^p}.$$

NB:  It might also be worth mentioning that this is precisely the same notation which is used to indicate the (Lebesgue) integral taken over some set, rather than an interval in $\mathbb{R}$.  This is natural, as sums can be thought of as integrals over a countable set with respect to the counting measure.
