Clarification of summation upper bound in $\sum_{p\text{ prime}} \frac{1}{2^p}$

Question

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.

Answer 1

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 $$ \sum_{\text{$p$ prime}} \frac{1}{2^p}.$$

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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.