This is trivial for $n=3$ and for $n=4$: $\frac14+\frac19+\frac1{25}+\frac1{30}<\frac14+\frac18+\frac1{16}+\frac1{16}=\frac12$, $\frac14+\frac19+\frac1{25}+\frac1{49}+\frac1{210}<0.25+0.12+0.04+0.02+0.01<\frac12$. For all $n>4$ we have $\frac1{p_1p_2\dots p_n}<\frac1{2\cdot 3\cdot 5\cdot 7\cdot 11}<0.0005$, so it is enough to prove that $S_n:=\sum_{k=1}^n p_k^{-2}<0.4995$ for $n>3$. Note that $S_6=\frac1{2^2}+\frac1{3^2}+\frac1{5^2}+\frac1{7^2}+\frac1{11^2}+\frac1{13^2}<0.437$, this proves the inequality for $n\le 6$. For $n>6$, setting $m=p_n^2$ we can estimate the rest as follows: $\sum_{k=7}^n p_k^{-2}<\sum_{i=17}^m\frac1{i^2}<$ $<\sum_{i=17}^m\frac1{i(i-1)}<\sum_{i=17}^m\left(\frac1{i-1}-\frac1i\right)=\frac1{16}-\frac1m<\frac1{16}$. Therefore $S_n<\frac1{16}+0.437=0.4995$ indeed. So we're done for all $n$.