Let's use Euler's notation that involves: $$\sum_{n=1}^{\infty}{\frac{1}{n}} = \ln{\infty}$$ And $$\sum_{k=1}^{\infty}{\frac{1}{p_k}} = \ln{\ln{\infty}}$$
Could we say that $$\sum_{k=1}^{\infty}{\frac{1}{p_{2k}}} = \frac{\ln{\ln{\infty}}}{2}$$ ? Why/why not?
Thank you.
If you want to be understood, use modern notation, not Euler's. I think what you're saying is
$$ \eqalign{\sum_{n=1}^N \dfrac{1}{n} &\sim \log(N) \ \text{as}\ N \to \infty \cr \sum_{n=1}^N \dfrac{1}{p_n} &\sim \log \log(N) \ \text{as}\ N \to \infty \cr}$$
Now since $p_{k+1} > p_k$ $$ \frac{1}{2} \sum_{n=2}^{2N+1} \dfrac{1}{p_n}\le \sum_{n=1}^N \dfrac{1}{p_{2n}} \le \frac{1}{2}\sum_{n=1}^{2N} \dfrac{1}{p_n}$$ and thus $$ \sum_{n=1}^N \dfrac{1}{p_{2n}} \sim \frac{\log \log(2N)}{2} \sim \frac{\log \log(N)}{2} \ \text{as}\ N \to \infty $$