Here is a question I have asked on Math Stack Exchange <https://math.stackexchange.com/questions/2290917/prime-numbers-property-mertens-theorem-related-sequence> , that I would like this community to address. Yes someone responded there, but it seems to have some errors. So: Merten's third theorem states that $$ \lim\_{n \to \infty} \log(p\_n) \prod\_{i=1}^n \left( 1 - \frac{1}{p\_i}\right) = e^{-\gamma}$$ This is a particularly interesting product since $\log(p\_n) \rightarrow \infty$ and $\prod\_{i=1}^n \left( 1 - \frac{1}{p\_i}\right) \rightarrow 0 $ hence Merten's theorem states a sort of balance between the two sequences ... Now obviously $$ \lim\_{n \to \infty} p\_n \prod\_{i=1}^n \left( 1 - \frac{1}{p\_i}\right) = \infty$$ but let $\alpha \in [0,1]$ and let $$ T\_{\alpha}(p\_n) = p\_n^{\alpha} \prod\_{i=1}^n \left( 1 - \frac{1}{p\_i^{\alpha}}\right)$$ and observe that $\lim\_{n \to \infty} T\_{\alpha = 0} (p\_n) = 0$ and $\lim\_{n \to \infty} T\_{\alpha = 1} (p\_n) = \infty$. My question is: is there an $\alpha^\* \in (0,1)$ such that $\lim\_{n \to \infty} T\_{\alpha^\* } (p\_n) = k$ with $k \in \mathbb{R}$? According to some of my numerical calculations, this is the case for $\alpha < 4/5$ and possibly above a little. The limit is $k = 0$ in these cases. But I am of course interested in a demonstration and if there is some $\alpha$ for which the limit is not zero but finite.



Prime numbers property. A Merten's third theorem like sequence
I don't think this question is of research level, but let me answer it. Using that the natural logarithm is a concave function, we have
$$ \log T\_\alpha(p\_n)=\alpha\log p\_n+\sum\_{i=1}^n\log\left(1-\frac{1}{p\_i^\alpha}\right)<\alpha\log p\_n-\sum\_{i=1}^n\frac{1}{p\_i^\alpha}.\tag{$\*$}$$
Using the Lebesgue-Stieltjes integral and then integration by parts, the sum on the right hand side equals
$$\sum\_{i=1}^n\frac{1}{p\_i^\alpha}=\int\_{2-}^{p\_n}\frac{d\pi(x)}{x^\alpha}=\frac{\pi(p\_n)}{p\_n^\alpha}+\alpha\int\_2^{p\_n}\frac{\pi(x)}{x^{\alpha+1}}\,dx.$$
Using the Prime Number Theorem and the fact that the logarithm grows slowly, the right hand side equals
\begin{align\*}
(1+o(1))\left\{\frac{p\_n^{1-\alpha}}{\log p\_n}+\alpha\int\_2^{p\_n}\frac{dx}{x^\alpha\log x}\right\}
&=(1+o(1))\left\{\frac{p\_n^{1-\alpha}}{\log p\_n}+\frac{\alpha}{\log p\_n}\int\_2^{p\_n}\frac{dx}{x^\alpha}\right\}\\
&=\frac{1+o(1)}{1-\alpha}\cdot\frac{p\_n^{1-\alpha}}{\log p\_n}.
\end{align\*}
That is, on the right hand side of $(\*)$, the sum always grows much faster than $\log p\_n$, whence 
$$\lim\_{n\to\infty}\log T\_\alpha(p\_n)=-\infty.$$
This means that $T\_\alpha(p\_n)$ tends to zero for any $\alpha\in(0,1)$.


Actually, we don't need the Prime Number Theorem to reach this conclusion, Chebyshev's bound $\pi(x)\gg x/\log x$ suffices for the purpose. Indeed, using this bound only, we can estimate the sum on the right hand side of $(\*)$ as follows, starting from the display below ($\*$):
$$\sum\_{i=1}^n\frac{1}{p\_i^\alpha}\gg\_\alpha\int\_2^{p\_n}\frac{\pi(x)}{x^{\alpha+1}}\,dx\gg \int\_2^{p\_n}\frac{dx}{x^\alpha\log x}\geq\frac{1}{\log p\_n}\int\_2^{p\_n}\frac{dx}{x^\alpha}\gg\_\alpha\frac{p\_n^{1-\alpha}}{\log p\_n}.$$


