We improve the bound to $8$. Lemma: $\sum_{k=1}^{2^n-1} \frac{1}{k}<n$ for $n \ge 2$. Proof: By induction on $n$. The base case $n=2$ is easy as $1+\frac{1}{2} + \frac{1}{3} = \frac{11}{6}<2$. Now assume $\sum_{k=1}^{2^n-1} \frac{1}{k}<n$ for some $n \ge 2$: \[\sum_{k=1}^{2^{n+1}-1} \frac{1}{k} = \sum_{k=1}^{2^n-1} \frac{1}{k} + \sum_{k=2^n}^{2^{n+1}-1} \frac{1}{k}<n+\sum_{k=2^n}^{2^{n+1}-1} \frac{1}{2^n}=n+1\]Which completes the proof. Let $S$ be the desired sum. Let $X$ be the set of all primes less than $2^{100}$, and also including $1$. Notice that numbers of the form $p_1p_2p_3p_4p_5p_6$, where $p_i \in X$ and $p_1 \le p_2 \le p_3 \le p_4 \le p_5 \le p_6$ are distinct integers in $[1,2,\cdots ,2^{600}-1]$. We get: \[\left( 1+S \right) ^6 = \left( \sum_{p \in X} \frac{1}{p} \right) ^6 = 6! \sum_{p \in X} \frac{1}{p_1p_2p_3p_4p_5p_6} < 6! \sum_{k=1}^{2^{600}-1} \frac{1}{k}<6! \cdot 600 = 720 \cdot 600 < 729 \cdot 729 = 3^{12}\]Thus $1+S<3^2$ so $S<8$.

Google Mertens's Second estimate lol. (It actually can be proven quickly, though not trivially, by elementary terms - no need to really rely on Partial Summation or the Prime Number Theorem.)

Verbatim from Romania 2015 10.4