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.
Since $-\log(1-x) = x+\mathcal{O}(x^2)$ for $\alpha > 1$, $\prod_{k=1}^\infty (1-n^{-\alpha})$ converges as well as $\prod_{k=1}^\infty (1-p_k^{-\alpha})= \frac{1}{\zeta(\alpha)}$
For $\alpha \in (1/2,1]$ : $\sum_{k=1}^\infty p_k^{-2\alpha}$ converges so that $$-\log \prod_{k=1}^K (1-p_k^{-\alpha}) =\underbrace{\mathcal{O}(1)+ \sum_{k=1}^K p_k^{-\alpha} \sim \sum_{n=2}^{p_K} \frac{n^{-\alpha}}{\log n}}_{\text{prime number theorem}} \sim \mathcal{O}(1)+C\frac{p_K^{1-\alpha}}{\log p_K}$$
and by the PNT again $p_K \sim K \log K$
For $\alpha > 0$ you can make the same reasoning $-\log \prod_{k=1}^K (1-p_k^{-\alpha}) \sim \sum_{k=1}^K p_k^{-\alpha} \sim \sum_{n=1}^{p_K} \frac{n^{-\alpha}}{\log n} \sim C\frac{p_K^{1-\alpha}}{\log K}$
Thus the answer is that $\lim_{K \to \infty} p_K^\alpha \prod_{k=1}^K (1-p_k^{-\alpha}) = 0$ for $\alpha \in \mathbb{C},\alpha \in (0,1)$ (or $\Re(\alpha) \in (0,1)$) otherwise it diverges
In number theory $\sum_p$ means summing over the prime, $\sum_n$ means summing over the integers. Let $\Re(\alpha) \in (0,1)$ so that everything $\to \infty$,
$L(N) = \sum_{ n=2}^N \frac{1}{\ln n}$ and $\pi(N) = \sum_{p \le N} 1$.
By the PNT $L(N) \sim \pi(N)$ so that $ L(N) = \pi(N)(1+o(1))$. Summing by parts
$$\sum_{n =1}^N \pi(n) (n^{-\alpha}-(n+1)^{-\alpha}) =\pi(N)(N^{-\alpha}-(N+1)^{-\alpha}) +\sum_{p < N} p^{-\alpha} $$
$$\sum_{n =1}^N L(n) (n^{-\alpha}-(n+1)^{-\alpha}) =L(N)(N^{-\alpha}-(N+1)^{-\alpha}) +\sum_{n < N} \frac{n^{-\alpha}}{\ln n} $$
where $N^{-\alpha}-(N+1)^{-\alpha} = \int_N^{N+1} \alpha x^{-\alpha-1}dx \sim \alpha N^{-\alpha-1}$ so that $L(N)(N^{-\alpha}-(N+1)^{-\alpha})$ is much smaller than the sum
and
$$\sum_{n =1}^N \pi(n) (n^{-\alpha}-(n+1)^{-\alpha})=\sum_{n =1}^N L(N)(1+o(1)) (n^{-\alpha}-(n+1)^{-\alpha}) \\ \sim \sum_{n =1}^N L(N)(n^{-\alpha}-(n+1)^{-\alpha})$$
Qed
$$\sum_{p < N} p^{-\alpha} \sim \sum_{2 \le n < N} \frac{n^{-\alpha}}{\ln n}$$
Finally $\frac{d}{d\alpha}\sum_{2 \le n < N}\frac{n^{-\alpha}}{\ln n}=-\sum_{2 \le n < N}n^{-\alpha} \sim \frac{N^{1-\alpha}-1}{1-\alpha}$ and
$$\sum_{p < N} p^{-\alpha} \sim \sum_{2 \le n < N}\frac{n^{-\alpha}}{\ln n} \sim \int_1^\alpha \frac{N^{1-a}-1}{a-1}da\sim \int_1^\alpha \frac{N^{1-a}-1}{\alpha-1}da = \frac{N^{1-\alpha}-2}{(1-\alpha)\ln N}\sim \frac{N^{1-\alpha}}{(1-\alpha)\ln N}$$
as claimed