[proofplan]
We prove convergence by reducing the Brun series to a dyadic estimate for the first prime in each twin-prime pair. The only sieve input is Brun's upper bound for the twin-prime counting function, namely that the number of primes $p \leq x$ with $p + 2$ prime is at most a constant multiple of $x(\log\log x)^2/(\log x)^2$. On each dyadic interval $[2^m,2^{m+1})$, this bound controls both the number of contributing primes and the size of each reciprocal term. The resulting majorant is a convergent numerical series, and the terms $1/(p+2)$ are dominated by $1/p$.
[/proofplan]
[step:Define the twin-prime counting function and record Brun's upper bound]
Let $T \subset \mathbb{P}$ denote the set
\begin{align*}
T := \{p \in \mathbb{P} : p + 2 \in \mathbb{P}\}.
\end{align*}
Define the twin-prime counting function
\begin{align*}
\pi_2 : [2,\infty) \to \mathbb{N} \cup \{0\}, \qquad \pi_2(x) := \#\{p \in T : p \leq x\}.
\end{align*}
We use [Brun's upper bound for twin primes](/theorems/7167): there exist constants $C_0 > 0$ and $x_0 \geq 3$ such that, for every $x \geq x_0$,
\begin{align*}
\pi_2(x) \leq C_0 \frac{x(\log\log x)^2}{(\log x)^2}.
\end{align*}
This sieve-theoretic estimate is the only external input of the proof.
[/step]
[step:Bound the reciprocal sum on each dyadic interval]
Choose an integer $m_0 \in \mathbb{N}$ such that $2^{m_0} \geq x_0$ and $m_0 \geq 2$. For each integer $m \geq m_0$, define the finite dyadic block
\begin{align*}
T_m := T \cap [2^m,2^{m+1}).
\end{align*}
If $p \in T_m$, then $p \geq 2^m$, so $1/p \leq 2^{-m}$. Therefore
\begin{align*}
\sum_{p \in T_m} \frac{1}{p} \leq 2^{-m}\#T_m.
\end{align*}
Since $T_m \subset \{p \in T : p \leq 2^{m+1}\}$, we have $\#T_m \leq \pi_2(2^{m+1})$. Brun's upper bound gives
\begin{align*}
\#T_m \leq C_0 \frac{2^{m+1}(\log\log 2^{m+1})^2}{(\log 2^{m+1})^2}.
\end{align*}
Combining these estimates,
\begin{align*}
\sum_{p \in T_m} \frac{1}{p} \leq 2C_0 \frac{(\log((m+1)\log 2))^2}{((m+1)\log 2)^2}.
\end{align*}
Thus there is a constant $C_1 > 0$, depending only on $C_0$ and the fixed number $\log 2$, such that for every $m \geq m_0$,
\begin{align*}
\sum_{p \in T_m} \frac{1}{p} \leq C_1 \frac{(\log(m+1))^2}{(m+1)^2}.
\end{align*}
[guided]
The purpose of the dyadic decomposition is to separate the two pieces of information we have: Brun's bound controls how many twin primes occur up to a scale, while the reciprocal $1/p$ is essentially constant on a dyadic scale.
Choose $m_0 \in \mathbb{N}$ with $2^{m_0} \geq x_0$ and $m_0 \geq 2$. For each $m \geq m_0$, define
\begin{align*}
T_m := T \cap [2^m,2^{m+1}).
\end{align*}
This is the set of first components of twin-prime pairs whose size is comparable to $2^m$. If $p \in T_m$, then $p \geq 2^m$, hence
\begin{align*}
\frac{1}{p} \leq \frac{1}{2^m}.
\end{align*}
Summing this bound over the finite set $T_m$ gives
\begin{align*}
\sum_{p \in T_m} \frac{1}{p} \leq 2^{-m}\#T_m.
\end{align*}
Now we estimate $\#T_m$. Since every element of $T_m$ is at most $2^{m+1}$, the inclusion
\begin{align*}
T_m \subset \{p \in T : p \leq 2^{m+1}\}
\end{align*}
implies
\begin{align*}
\#T_m \leq \pi_2(2^{m+1}).
\end{align*}
Because $m \geq m_0$, we have $2^{m+1} \geq x_0$, so Brun's upper bound applies at $x = 2^{m+1}$. Therefore
\begin{align*}
\#T_m \leq C_0 \frac{2^{m+1}(\log\log 2^{m+1})^2}{(\log 2^{m+1})^2}.
\end{align*}
Multiplying by $2^{-m}$ gives
\begin{align*}
\sum_{p \in T_m} \frac{1}{p} \leq 2C_0 \frac{(\log\log 2^{m+1})^2}{(\log 2^{m+1})^2}.
\end{align*}
Since $\log 2^{m+1} = (m+1)\log 2$ and $\log\log 2^{m+1} = \log((m+1)\log 2)$, this becomes
\begin{align*}
\sum_{p \in T_m} \frac{1}{p} \leq 2C_0 \frac{(\log((m+1)\log 2))^2}{((m+1)\log 2)^2}.
\end{align*}
The factor $\log 2$ is fixed. Hence there is a constant $C_1 > 0$, depending only on $C_0$ and $\log 2$, such that
\begin{align*}
\sum_{p \in T_m} \frac{1}{p} \leq C_1 \frac{(\log(m+1))^2}{(m+1)^2}
\end{align*}
for all $m \geq m_0$.
[/guided]
[/step]
[step:Compare the dyadic majorant with a convergent numerical series]
We claim that
\begin{align*}
\sum_{m=m_0}^{\infty} \frac{(\log(m+1))^2}{(m+1)^2}
\end{align*}
converges. Since
\begin{align*}
\lim_{m \to \infty} \frac{(\log(m+1))^2}{(m+1)^{1/2}} = 0,
\end{align*}
there exists $m_1 \geq m_0$ such that
\begin{align*}
(\log(m+1))^2 \leq (m+1)^{1/2}
\end{align*}
for every $m \geq m_1$. Hence, for $m \geq m_1$,
\begin{align*}
\frac{(\log(m+1))^2}{(m+1)^2} \leq \frac{1}{(m+1)^{3/2}}.
\end{align*}
The $p$-series $\sum_{m=m_1}^{\infty}(m+1)^{-3/2}$ converges, and the finitely many terms with $m_0 \leq m < m_1$ do not affect convergence. Therefore the dyadic majorant converges.
[/step]
[step:Sum over all dyadic intervals to prove convergence of the first reciprocal series]
The sets $T_m$ for $m \geq m_0$ are pairwise disjoint, and their union is $T \cap [2^{m_0},\infty)$. For each integer $M \geq m_0$, finite additivity of sums over the disjoint finite sets $T_m$ gives
\begin{align*}
\sum_{m=m_0}^{M} \sum_{p \in T_m} \frac{1}{p} \leq C_1 \sum_{m=m_0}^{M} \frac{(\log(m+1))^2}{(m+1)^2}.
\end{align*}
Both sides are increasing in $M$ because all summands are non-negative. Taking the supremum over $M \geq m_0$, which is the monotone definition of a non-negative series, yields
\begin{align*}
\sum_{\{p \in T : p \geq 2^{m_0}\}} \frac{1}{p} = \sum_{m=m_0}^{\infty} \sum_{p \in T_m} \frac{1}{p} \leq C_1 \sum_{m=m_0}^{\infty} \frac{(\log(m+1))^2}{(m+1)^2}.
\end{align*}
The series on the right converges by the previous step. The remaining set $T \cap [2,2^{m_0})$ is finite, so
\begin{align*}
\sum_{\{p \in T : p < 2^{m_0}\}} \frac{1}{p}
\end{align*}
is a finite sum. Hence
\begin{align*}
\sum_{p \in T} \frac{1}{p}
\end{align*}
converges.
[/step]
[step:Dominate the second reciprocal series by the first]
For every $p \in T$, we have $p \geq 2$ and $p+2 > p$, so
\begin{align*}
0 < \frac{1}{p+2} \leq \frac{1}{p}.
\end{align*}
If $F \subset T$ is finite, summing the pointwise inequality over $F$ gives
\begin{align*}
0 \leq \sum_{p \in F} \frac{1}{p+2} \leq \sum_{p \in F} \frac{1}{p}.
\end{align*}
Taking the supremum over all finite subsets $F \subset T$ and using the monotone definition of a non-negative series gives
\begin{align*}
0 \leq \sum_{p \in T} \frac{1}{p+2} \leq \sum_{p \in T} \frac{1}{p}.
\end{align*}
The right-hand series converges by the previous step, so the [comparison test](/theorems/173) for non-negative series implies that $\sum_{p \in T} 1/(p+2)$ converges. Consequently
\begin{align*}
\sum_{\{p \in \mathbb{P} : p+2 \in \mathbb{P}\}} \left(\frac{1}{p}+\frac{1}{p+2}\right) = \sum_{p \in T}\frac{1}{p} + \sum_{p \in T}\frac{1}{p+2}
\end{align*}
is the sum of two convergent non-negative series. This proves the convergence of the twin-prime reciprocal series.
[/step]
