Asymptotic behaviour of $\frac{p_n-2}{p_n}$ I am interested in how fast this prime product approaches to 0:
$$
\prod_{n=2}^{N}\frac{p_n - 2}{p_n}
$$
Where $p_n$ is the nth prime number.
Numerical computation suggests that for $N > 1312$, $k = 0.5$
we have the following upper bound:
$$
\prod_{n=2}^{N}\frac{p_n - 2}{p_n} > \frac{1}{p_N^{k}}
$$
Can anyone prove this?
 A: We know that $n \ln n < p_n $ for all $n\geq 1$ and you are looking for lower bound so $\prod \limits_{n=2}^{N} (1-\frac{2}{p_n}) = \frac{1}{5} \prod \limits_{n=4}^{N} (1-\frac{2}{p_n}) \geq 0.2 \prod \limits_{n=4}^{N} (1-\frac{2}{n \ln n}) = 0.2 e^{\sum \limits_{n=4}^{N} \ln(1-\frac{2}{n \ln n})}$ and since $\ln(1-\frac{2}{t}) > -\frac{3}{t}$ for all $t>3.44$ and that $n \ln n > 3.44$ for all $n\geq 4$.
So we are left with $0.2 e^{-3 \sum \limits_{n=4}^{N} \frac{1}{n \ln n}}$
$\frac{1}{n \ln n}$ is monotonic deceasing function so we can bound the summation by integral, and we get that $0.2 e^{-3 \sum \limits_{n=4}^{N} \frac{1}{n \ln n}} \geq 0.2 e^{-3 \int \limits_{4}^{N} \frac{1}{n \ln n}dn}$
and $0.2 e^{-3 \int \limits_{4}^{N} \frac{1}{n \ln n}dn} = 0.2 e^{-3(\ln \ln N-\ln \ln 4)}$ since $\int \frac{1}{n \ln n} dn = \ln \ln n+C$
Now we are left with $0.2 e^{-3(\ln \ln N-\ln \ln 4)} = \frac{\ln^3(4)}{5 \ln^3(N)} \approx \frac{0.53284}{\ln ^3 N} > \frac{1}{p_N^{0.5}}$ for all $N>10^7$
Actually a very good approach using Dusart result about the sum  $ \ln \ln x +B -\frac{1}{2 \ln^2 x}<\sum \limits_{p \leq x} \frac{1}{p} < \ln \ln x +B +\frac{1}{2 \ln^2 x}$ for $x>300$ where $B \approx 0.2614972128476427$ is Meissel–Mertens constant, one can bound your product by :
$\frac{0.814886}{\ln^2 p_N} <\prod \limits_{n=2}^{N} (1-\frac{2}{p_n}) < \frac{0.843639}{\ln^2 p_N}$
And the exact value is $c = \lim \limits_{N \to \infty} \ln^2 p_N \prod \limits_{n=2}^{N} (1-\frac{2}{p_n}) \approx 0.832417$
in this case $\prod \limits_{n=2}^{N} (1-\frac{2}{p_n}) > \frac{0.814886}{\ln^2 p_N} > \frac{1}{p_N^{0.5}}$ for all $N>1388$ , by computer checking for all $1388 \geq N \geq 1312$ one conclude that its true for all $N>1312$.
