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The Selberg upper-bound sieve replaces the indicator function of a sifted set by the square of a divisor sum. If the real Selberg sieve weights satisfy ${\lambda }_1=1$, thenWhen $F(n)$ has no forbidden prime factors, the divisor sum is exactly one; otherwise its square is still nonnegative. Taking the weights to vanish for $d>z$ controls the error in summing this majorant. The main term is a positive quadratic form that can be diagonalized and minimized explicitly. This optimization, together with a bound for the remainder, is the central idea of the Selberg sieve.

For twin primes, take $F(n)=n(n+2)$ and let $\rho (d)$ be the number of its roots modulo a squarefree integer $d$. The Chinese remainder theorem makes $\rho$ a multiplicative arithmetic function, with $\rho (2)=1$ and $\rho (p)=2$ for odd primes. Thus the polynomial root density in a sieve is $g(d)=\rho (d)/d$, andSumming the majorant giveswhere $Q(\lambda )={\sum }_{d,e\le z}{\lambda }_d{\lambda }_{e}g([d,e])$, with all indices squarefree. Here $\rho ([d,e])\le \rho (d)\rho (e)$ controls the error.

We carry out the Selberg sieve diagonalization. PutFor squarefree integers $d,e$, multiplicativity givesThe identity behind the first equality is ${\prod }_{p\mid (d,e)}[1+h(p)]=1/g((d,e))$. Möbius inversion gives ${\lambda }_1={\sum }_r\mu (r)y_r=1$. The Cauchy-Schwarz inequality therefore impliesDefine $G(z)={\sum }_{r\le z,r\text{ squarefree}}1/h(r)$. Equality is attained at $y_r=\mu (r)/(h(r)G(z))$, giving $\min Q=1/G(z)$. Inverting again, the optimizing Selberg sieve weights areIn particular ${\lambda }_1=1$. Since $G_d(z/d)\le G(z)$,The last inequality holds separately at $p=2$, $p=3$, and $p\ge 5$. Together with $\rho (d)\le d$, this crude bound already makes the remainder $O(z^6)$, because ${\sum }_{d\le z}{d}^2=O(z^3)$.

We also need a lower bound for $G(z)$; we prove it rather than presuppose a sieve asymptotic. For odd squarefree integers $r$,where $\omega (r)$ counts distinct prime factors. Put $y=\sqrt{z}$ and $L(y)={\sum }_{a\le y,a\text{ odd}}1/a=\frac{1}{2}\log y+O(1)$. The total weight $1/(ab)$ of odd pairs $a,b\le y$ is $L(y{)}^2$. Exclude pairs for which $a$ has a square prime factor, $b$ has a square prime factor, or $a,b$ have a common prime factor. For each of these three categories the total weight is at most $L(y{)}^2{\sum }_{p\text{ odd}}{p}^{-2}$: for example, the weighted count of odd multiples of $p^2$ is $p^{-2}L(y/p^2)\le p^{-2}L(y)$, and the common-factor category uses $p^{-1}L(y/p)$ for each variable. Moreover,Hence the remaining odd coprime squarefree pairs have weight at least $L(y{)}^2/6$. Their products $r=ab\le z$ are squarefree, and $2^{\omega (r)}$ counts all ordered coprime factorizations of $r$. This proves the elementary lower bound for the twin-prime sieve denominator:

If $p>z$ and both $p,p+2$ are primes, $F(p)$ survives the sieve. The at most $z$ smaller values are the only exceptions. Thus, with $T(x)=\#\{p\le x:p,p+2\text{ prime}\}$ and $z=x^{1/10}$, the twin-prime upper bound from a quadratic sieve becomesFinally partial summation givesThe integral over $[3,\infty )$ is bounded by a constant times ${\int }_3^{\infty }dt/(t(\log t{)}^2)<\infty$. The nonnegative partial sums are therefore bounded and increasing. Thus the reciprocal series over the smaller members of twin-prime pairs converges, a form of Brun's theorem. The upper bound does not assert that infinitely many twin primes exist.