In the proof of the Hardy–Ramanujan theorem by Paul Turan, at the end of the proof, it is stated that
If, instead of (4), we use the more precise but also elementary relation $$\sum_{p_i \leq M} \frac{1}{p_i} = \log \log M + a + o(1) $$ where $a$ is a constant, we obtain in the same way the results $$R(N) = N \log \log N + o(N \log \log N)$$
I tried to follow the proof but I can't get to the desired conclusion.
In the proof of Lemma 4, it uses that $$\left( \sum_{p_i \leq \sqrt{N}} \frac{1}{p_i} \right)^2 < \sum_{p_i p_h < N} \frac{1}{p_i p_h} < \left( \sum_{p_i \leq N} \frac{1}{p_i} \right)^2$$
If I use the Merten's Second Theorem, $$\sum_{p \leq x} \frac{1}{p} = \log \log x + B_1 + o(1)$$ where $B_1$ is the Mertens constant, I get:
\begin{align*} \left( \sum_{p_i \leq N} \frac{1}{p_i} \right)^2 & = \left( \log \log N + B_1 + o(1) \right)^2 \\ & = \left( \log\log N \right)^2 + 2 B_1 \log \log N + B_1^2 + o(\log \log N) \\ & = \left( \log\log N \right)^2 + 2 B_1 \log \log N + o(\log \log N) \end{align*}
\begin{align*} \left( \sum_{p \leq \sqrt N} \frac{1}{p} \right)^2 & = \left( \log\log \sqrt N \right)^2 + 2 B_1 \log \log \sqrt N + o(\log \log \sqrt N) \\ & = \left( \log\log N - \log 2 \right)^2 + 2 B_1 \left( \log\log N - \log 2 \right) + o(\log\log N - \log 2) \\ & = \left( \log \log N \right)^2 - 2 \log 2 \left( \log \log N \right) + 2B_1 \log \log N + o(\log \log N) \\ & = \left( \log \log N \right)^2 + (2 B_1 - 2\log 2) \log \log N + o(\log \log N) \end{align*}
Using this, I can only say that there exists a constant $c$ such that: $$\sum_{p_i p_h \leq N} \frac{1}{p_i p_h} = \left( \log \log N \right)^2 + \left( 2B_1 - c \right) \log \log N + o(\log \log N)$$ where $0 < c < 2 \log 2$.
Then, following the proof, we can get
$$\sum_{n = 1}^{N} V(N)^2 = N(\log \log N)^2 + (2 B_1 - c + 1) N \log \log N + o(N \log \log N)$$
Also, we use the following result: $$\sum_{n \leq x} V(n) = x \log \log x + B_1 x + o(x)$$
Lastly,
\begin{align*} R(N) & = \sum_{n = 1}^N (V(N) - \log \log N)^2 \\ & = \sum_{n = 1}^N V(n)^2 - 2 \log \log N \sum_{n = 1}^{N} V(n) + N(\log \log N)^2 \\ & = N(\log \log N)^2 + (2 B_1 - c + 1) N \log \log N + o(N \log \log N) - 2 \log \log N \left( N \log \log N + B_1 N + o(1) \right) + N(\log \log N)^2 \\ & = (-c + 1) N \log \log N + o(N \log \log N) \\ \end{align*}
It seems like that I can do something better to eliminate the $c$, but I am not aware of a way to do so. Could you give me some suggestions?