Limit of Sequence of unusual Prime Product

Let $p_n$ be the nth prime and $p_L$ be closest to its square root: \begin{equation} p_L^2 \approx p_n \approx x \end{equation}

Let $\sigma \in Z^+$ be a positive integer constant. Define the average slope as \begin{equation} M_{n} = \prod_{\sigma < p_i \leq p_n} \left( \frac{p_i - \sigma}{p_i} \right) \end{equation}

Asymptotically the average slope becomes \begin{equation} M_{n} \sim \frac{K(\sigma)}{e^{\sigma \gamma} (\ln p_n)^\sigma} \quad n \rightarrow \infty \end{equation} where $\gamma$ is the Euler-Mascheroni Constant and \begin{equation} K(\sigma) = \sum_{p_i \leq \sigma} \frac{\sigma}{p_i} \end{equation}

Now define the square weighted slope as \begin{equation} S_{n} = \sum_{p_L < p_i \leq p_n} \frac{p_i^2 - p_{i-1}^2}{p_n^2 - p_L^2} M_{i-1} = \frac{(p_{L+1}^2 - p_{L}^2) M_{L} + \ldots + (p_{n}^2 - p_{n-1}^2) M_{n-1}}{p_n^2 - p_L^2} \end{equation}

Asympototically, \begin{equation} S_{n} \sim \frac{C(\sigma)}{(\ln p_n)^\sigma} \quad n \rightarrow \infty \end{equation} My question is: what is $C(\sigma)$? Is $C(\sigma) = K(\sigma)$? If $C(1) = K(1) = 1$, then I have an exciting new proof of the Prime Number Theorem.

More information can be found at: http://www.ugcs.caltech.edu/~kel/MPP/GammaSquares.pdf Thank you.