We need an upper bound on \(V^+(z)\) and a lower bound on \(V^-(z)\text{;}\) we get both of these by getting an upper bound on \(V_n(z)\text{.}\) First, let us simplify the sum by relaxing the summation conditions. We claim that for any tuple \(p_1, \dots, p_n\) appearing in the sum defining \(V_n(z)\text{,}\) and any \(m \lt n\text{,}\)
\begin{equation}
p_1 \cdots p_{m-1} p_m^\beta \lt y\text{.}\tag{12.4.1}
\end{equation}
Namely, if \(m \equiv n \pmod{2}\text{,}\) we have the stronger inequality
\begin{equation*}
p_1 \cdots p_{m-1} p_m^{1+\beta} \lt y\text{.}
\end{equation*}
If \(m > 1\) and \(m \not\equiv n \pmod{2}\text{,}\) we have
\begin{equation*}
p_1 \cdots p_{m-1} p_m^\beta \lt p_1 \cdots p_{m-2} p_{m-1}^{1+\beta} \lt y\text{.}
\end{equation*}
Finally, if \(m = 1\) and \(m \not\equiv n \pmod{2}\text{,}\) we have
\begin{equation*}
p_1^\beta \lt z^\beta = y^{\beta/s} \leq y\text{.}
\end{equation*}
From
(12.4.1), we deduce by induction on
\(m\) that
\begin{equation*}
p_1 \cdots p_m \lt y^{1 - (1 - \beta^{-1})^m} \qquad (m=1,\dots,n-1)\text{.}
\end{equation*}
In particular,
\begin{equation*}
p_n \geq (y/(p_1\cdots p_{n-1}))^{1/(\beta+1)}
\geq y^{\frac{1}{\beta+1} (1 - \beta^{-1})^{n-1}}
\geq y^{\frac{1}{\beta} (1 - \beta^{-1})^{n}}
\geq z_n
\end{equation*}
if we put
\begin{equation*}
z_n = z^{(1-\beta^{-1})^n}\text{.}
\end{equation*}
We will now retain only the conditions \(z > p_1 >\cdots > p_n \geq z_n\) on the primes, which will make the sum bigger because every summand is nonnegative. That is,
\begin{align*}
V_n(z) \amp\leq \sum_{z>p_1>\cdots>p_n \geq z_n} g(p_1\cdots p_n) V(p_n)\\
\amp\leq \frac{1}{n!} V(z_n) \left( \sum_{z_n \leq p \lt z} g(p) \right)^n\\
\amp\leq \frac{1}{n!} V(z_n) \left( \log \frac{V(z_n)}{V(z)} \right)^n\text{.}
\end{align*}
Here is where we need the assumption
(12.3.1) about the sieve dimension. It implies
\begin{equation*}
\frac{V(z_n)}{V(z)} \leq K (1 + (\beta-1)^{-1})^{\kappa n} \lt K e^{n/b}
\end{equation*}
for \(\beta = \kappa b + 1\) (using the bound \(1+x \leq e^x\) for \(x = (\beta-1)^{-1} = 1/(\kappa b)\)), which gives us
\begin{align*}
V_n(z) \amp\lt \frac{K}{n!} \left( \frac{n}{b} + \log K \right)^n e^{n/b} V(z)\\
\amp\leq \frac{K}{n!} \left( \frac{n}{b} e^{1/b} \right)^n K^b V(z)
\end{align*}
(using the bound \(1+x \leq e^x\) for \(x = b (\log K)/n\)). Since \(n! \geq e (n/e)^n\) (by taking logs and comparing integrals), we obtain
\begin{equation*}
V_n(z) \lt e^{-1} a^n K^{b+1} V(z)
\end{equation*}
for \(a = b^{-1} e^{1+b^{-1}}\text{.}\)
To conclude, we clean things up a bit. Remember that we were at liberty to choose \(\beta > 1\text{,}\) which is equivalent to choosing \(b > 0\text{.}\) By taking \(b\) sufficiently large, we can force \(a \lt 1\text{;}\) for instance, we could take \(b = 9\) to get \(a \lt e^{-1}\text{.}\) Note also that because
\begin{equation*}
p_1 \gt \cdots \gt p_n \geq
y_n = (y/(p_1 \cdots p_n))^{1/\beta}\text{,}
\end{equation*}
we have \(p_1^{n+\beta} > y\text{.}\) Since we also have \(p_1 \lt z = y^{1/s}\text{,}\) we deduce that \(V_n(z) = 0\) unless \(n + \beta > s\text{.}\) Therefore
\begin{equation*}
\sum_{n>0} V_n(z) = \sum_{n>s-\beta} V_n(z) \lt \frac{a^{s-\beta}}{e(1-a)}
K^{b+1} V(z)\text{.}
\end{equation*}
To conclude, we have the following bound.
