I have revised my original post. The questions I asked there were not well-put or even thought through. I don't want to delete, however, since some of the comments may be of interest to other MO users.

It is well known that the sums of the reciprocals of the primes diverges. But suppose we decompose the set of prime numbers, call this set $P$, into disjoint infinite sets $\langle A_\alpha\rangle$ in some way to guarantee that each member $A_\alpha$ of the partition has the following "nice" property: $$\Sigma_{p\in A_{\alpha}} \frac{1}{p} \text{ converges}.$$

There are by now several interesting partitions in the comments and answers below. Here I include two of my own attempts at such a partition.

Attempt 1: Let $\langle \epsilon_i: i<\omega\rangle$ be a sequence of positive real numbers be such that $\Sigma_i \epsilon_i$ diverges (goes to infinity). Suppose you want an infinite $X\subseteq P$ such that $$\Sigma_{p\in X}\frac{1}{p} <\epsilon_1.$$ Then inductively define $X_1$ as follows: Pick $p_1\in P$ least such that $\frac{1}{p}<\epsilon_1$. Continuing, suppose $p_1,p_2,\dots,p_n$ have been chosen such that $$\Sigma^n_1\frac{1}{p_i}<\epsilon.$$ Then let $p_{n+1}$ be the least such that $$\Sigma^{n+1}_1\frac{1}{p_i}<\epsilon_1.$$

It seems like we could continue this strategy to produce a partition with the desired property by setting $P=P_1$, and $P_2=P_1\setminus X_1$. Then repeat the process described above. If this stage goes all the way through, define $P_3$, etc. It seems that all the $P_n$s will be defined and each will, by construction converge.

But this sort of strategy seems to depend on the paramater sequence $\langle\epsilon_i\rangle$ in an essential way. It may be constructive, but the strategy doesn't really give me a concrete or satisfying (or even really explicit) definition for the partition.

Attempt 2: I proposed the following partial strategy (based on the Green-Tao Theorem). Index the primes according to their natural order (i.e., $p_1 =2, p_2 =3, p_3=5$, etc.). Let $P=A_1\cup B_1$ where $$n\in A_1\Longleftrightarrow n \text{ has a prime index},$$ and $n\in B_1$ otherwise. Since $B$ has positive upper density, it must contain arbitrarily long arithmetic progressions by the Tao-Green Theorem (see the related question Extension of Tao-Green Theorem). I'm not sure if this means $$\Sigma_{p\in B_1} \text{ diverges }$$ but I think it does.

Also, Ben Green comments there that $A$ must contain arbitrarily long arithmetic progressions since the density of this set is roughly $1/\log^2(N)$ (though this may still be conjecture, I'm not sure).

But suppose we attempt to destroy each such arithmetic progression by decomposing both $A_1$ and $B_1$ in the following way: Re-index both $A_1$ and $B_1$ according to their natural order and partition each by placing an $n$ in one set of the partition if it is indexed by a prime, and placing it in the other set in the partition if it indexed by a composite.

If we repeat the construction $k$ times (for both $A_1$ and $B_1$ and all subsequently formed subsets) we must surely eliminate all arithmetic progressions with common difference $k$. But perhaps there are still arithmetic progressions with common difference $k'>k$. So we continue the iteration, creating a binary tree of height $\omega$, the leaves of which union to all of $P$.

It is still not clear to me that this partition will have the property I requested above. Certainly there are several nice examples in the comments and answers below.

I have revised my original post. The questions I asked there were not well-put or even thought through. I don't want to delete, however, since some of the comments may be of interest to other MO users.

It is well known that the sums of the reciprocals of the primes diverges. But suppose we decompose the set of prime numbers, call this set $P$, into disjoint infinite sets $\langle A_\alpha\rangle$ in some way to guarantee that each member $A_\alpha$ of the partition has the following "nice" property: $$\Sigma_{p\in A_{\alpha}} \frac{1}{p} \text{ converges}.$$

There are by now several interesting partitions in the comments and answers below. Here I include two of my own attempts at such a partition.

Attempt 1: Let $\langle \epsilon_i: i<\omega\rangle$ be a sequence of positive real numbers be such that $\Sigma_i \epsilon_i$ diverges (goes to infinity). Suppose you want an infinite $X\subseteq P$ such that $$\Sigma_{p\in X}\frac{1}{p} <\epsilon_1.$$ Then inductively define $X_1$ as follows: Pick $p_1\in P$ least such that $\frac{1}{p}<\epsilon_1$. Continuing, suppose $p_1,p_2,\dots,p_n$ have been chosen such that $$\Sigma^n_1\frac{1}{p_i}<\epsilon.$$ Then let $p_{n+1}$ be the least such that $$\Sigma^{n+1}_1\frac{1}{p_i}<\epsilon_1.$$

It seems like we could continue this strategy to produce a partition with the desired property by setting $P=P_1$, and $P_2=P_1\setminus X_1$. Then repeat the process described above. If this stage goes all the way through, define $P_3$, etc. It seems that all the $P_n$s will be defined and each will, by construction converge.

But this sort of strategy seems to depend on the paramater sequence $\langle\epsilon_i\rangle$ in an essential way. It may be constructive, but the strategy doesn't really give me a concrete or satisfying (or even really explicit) definition for the partition.

Attempt 2: I proposed the following partial strategy (based on the Green-Tao Theorem). Index the primes according to their natural order (i.e., $p_1 =2, p_2 =3, p_3=5$, etc.). Let $P=A_1\cup B_1$ where $$n\in A_1\Longleftrightarrow n \text{ has a prime index},$$ and $n\in B_1$ otherwise. Since $B$ has positive upper density, it must contain arbitrarily long arithmetic progressions by the Tao-Green Theorem (see the related question Extension of Tao-Green Theorem). I'm not sure if this means $$\Sigma_{p\in B_1} \text{ diverges }$$ but I think it does.

Also, Ben Green comments there that $A$ must contain arbitrarily long arithmetic progressions since the density of this set is roughly $1/\log^2(N)$ (though this may still be conjecture, I'm not sure).

But suppose we attempt to destroy each such arithmetic progression by decomposing both $A_1$ and $B_1$ in the following way: Re-index both $A_1$ and $B_1$ according to their natural order and partition each by placing an $n$ in one set of the partition if it is indexed by a prime, and placing it in the other set in the partition if it indexed by a composite.

If we repeat the construction $k$ times (for both $A_1$ and $B_1$ and all subsequently formed subsets) we must surely eliminate all arithmetic progressions with common difference $k$. But perhaps there are still arithmetic progressions with common difference $k'>k$. So we continue the iteration, creating a binary tree of height $\omega$, the leaves of which union to all of $P$.

It is still not clear to me that this partition will have the property I requested above. Certainly there are several nice examples in the comments and answers below.

It is well known that the sums of the reciprocals of the primes diverges. But suppose we decompose the set of prime numbers, call this set $P$, into disjoint infinite sets $\langle A_\alpha: \alpha<2^{\aleph_0}\rangle$ (I picked $2^{\aleph_0}$ since it is the highest conceivable bound for such a partition) in some way to guarantee that each member $A_\alpha$ of the partition has the following "nice" property: $$\Sigma_{p\in A_{\alpha}} \frac{1}{p} \text{ converges}.$$

Preliminary Question: Is such a partition even possible?

This seems plausible: Let $\langle \epsilon_i: i<\omega\rangle$ be a sequence of positive real numbers be such that $\Sigma_i \epsilon_i$ diverges (goes to infinity). Suppose you want an infinite $X\subseteq P$ such that $$\Sigma_{p\in X}\frac{1}{p} <\epsilon_1.$$ Then inductively define $X_1$ as follows: Pick $p_1\in P$ least such that $\frac{1}{p}<\epsilon_1$. Continuing, suppose $p_1,p_2,\dots,p_n$ have been chosen such that $$\Sigma^n_1\frac{1}{p_i}<\epsilon.$$ Then let $p_{n+1}$ be the least such that $$\Sigma^{n+1}_1\frac{1}{p_i}<\epsilon_1.$$

It seems like we could continue this strategy to produce a countable partition with the desired property by setting $P=P_1$, and $P_2=P_1\setminus X_1$. Then repeat the process described above. If this stage goes all the way through, define $P_3$, etc. It seems that all the $P_n$s will be defined and each will, by construction converge.

Question 1: Is there a partition of $P$ that is "parameter free"?

I proposed the following partial strategy (based on the Green-Tao Theorem). Index the primes according to their natural order (i.e., $p_1 =2, p_2 =3, p_3=5$, etc.). Let $P=A_1\cup B_1$ where $$n\in A_1\Longleftrightarrow n \text{ has a prime index},$$ and $n\in B_1$ otherwise. Since $B$ has positive upper density, it must contain arbitrarily long arithmetic progressions by the Tao-Green Theorem (see the related question Extension of Tao-Green Theorem). I'm not sure if this means $$\Sigma_{p\in B_1} \text{ diverges }$$ but I think it does.

If we repeat the construction $k$ times (for both $A_1$ and $B_1$ and all subsequently formed subsets) we must surely eliminate all arithmetic progressions with common difference $k$. But perhaps there are still arithmetic progressions with common difference $k'>k$. So we continue the iteration to countable length, creating a binary tree of height $\omega$, the leaves of which union to all of $P$.

Question 2: Can the partial strategy proposed above be extended beyond up through stage $\omega$?

This also seems plausible, since at any finite stage $m$ we will have a set of primes with composite index (which therefore has positive upper density in the set of primes). The sums of reciprocals of this set must diverge (assuming $B_1$ diverges (I believe)).

If the answer to question 2 is yes, then we can continue the construction through $\omega$. And this observation invites

Question 3: Could this construction halt at some stage $\beta<2^{\aleph_0}$?

My guess is that it does so at $\omega_1^{CK}$, the Church-Kleene ordinal (based on the apparently constructive nature of the strategy).

If the above program can be carried out, then

Question 4: Are there other effective/constructive/computable strategies that give a "nice" partition of $P$?

As a result of the very few comments below, I think questions 1, 3, and 4 are moot. I think I have also answered (though I confess I'm not totally convinced yet) question 2. I would like to amend my questions to the following:

Revised Question Does the strategy I proposed (using a re-indexing) end before $\omega_1^{CK}$? And if so, at which ordinal?

I have revised my original post. The questions I asked there were not well-put or even thought through. I don't want to delete, however, since some of the comments may be of interest to other MO users.

It is well known that the sums of the reciprocals of the primes diverges. But suppose we decompose the set of prime numbers, call this set $P$, into disjoint infinite sets $\langle A_\alpha\rangle$ in some way to guarantee that each member $A_\alpha$ of the partition has the following "nice" property: $$\Sigma_{p\in A_{\alpha}} \frac{1}{p} \text{ converges}.$$

There are by now several interesting partitions in the comments and answers below. Here I include two of my own attempts at such a partition.

Attempt 1: Let $\langle \epsilon_i: i<\omega\rangle$ be a sequence of positive real numbers be such that $\Sigma_i \epsilon_i$ diverges (goes to infinity). Suppose you want an infinite $X\subseteq P$ such that $$\Sigma_{p\in X}\frac{1}{p} <\epsilon_1.$$ Then inductively define $X_1$ as follows: Pick $p_1\in P$ least such that $\frac{1}{p}<\epsilon_1$. Continuing, suppose $p_1,p_2,\dots,p_n$ have been chosen such that $$\Sigma^n_1\frac{1}{p_i}<\epsilon.$$ Then let $p_{n+1}$ be the least such that $$\Sigma^{n+1}_1\frac{1}{p_i}<\epsilon_1.$$

It seems like we could continue this strategy to produce a partition with the desired property by setting $P=P_1$, and $P_2=P_1\setminus X_1$. Then repeat the process described above. If this stage goes all the way through, define $P_3$, etc. It seems that all the $P_n$s will be defined and each will, by construction converge.

Attempt 2: I proposed the following partial strategy (based on the Green-Tao Theorem). Index the primes according to their natural order (i.e., $p_1 =2, p_2 =3, p_3=5$, etc.). Let $P=A_1\cup B_1$ where $$n\in A_1\Longleftrightarrow n \text{ has a prime index},$$ and $n\in B_1$ otherwise. Since $B$ has positive upper density, it must contain arbitrarily long arithmetic progressions by the Tao-Green Theorem (see the related question Extension of Tao-Green Theorem). I'm not sure if this means $$\Sigma_{p\in B_1} \text{ diverges }$$ but I think it does.

If we repeat the construction $k$ times (for both $A_1$ and $B_1$ and all subsequently formed subsets) we must surely eliminate all arithmetic progressions with common difference $k$. But perhaps there are still arithmetic progressions with common difference $k'>k$. So we continue the iteration, creating a binary tree of height $\omega$, the leaves of which union to all of $P$.

It is still not clear to me that this partition will have the property I requested above. Certainly there are several nice examples in the comments and answers below.

As a result of the very few comments below, I think questions 1, 3, and 4 are moot. I think I have also answered (though I confess I'm not totally convinced yet) question 2. I would like to amend my questions to the following:

Revised Question Does the strategy I proposed (using a re-indexing) end before $\omega_1^{CK}$? And if so, at which ordinal?