Signed sums of prime reciprocals

One approach to finding non-smooth numbers (which, assuming a factoring oracle, would lead to progress on the finding primes problem) is as follows.

Suppose we wish to show that at least one integer in [math]\displaystyle{ [n,n+\log n] }[/math] contains a prime factor larger than, say, [math]\displaystyle{ \log^{100} n }[/math] (thus breaking the square root barrier). We suppose this is not the case and obtain a contradiction.

It is then plausible that most integers in this interval contain about [math]\displaystyle{ \log^{1-o(1)} n }[/math] or so factors between, say, [math]\displaystyle{ \log^{10} n }[/math] and [math]\displaystyle{ \log^{100} n }[/math]. (Using the W-trick, one can probably eliminate primes much less than log n from consideration.) In particular, we get a set [math]\displaystyle{ p_1,\ldots,p_k }[/math] of about [math]\displaystyle{ \log^{2-o(1)} n }[/math] primes in [math]\displaystyle{ [\log^{10} n, \log^{100} n] }[/math] which each divide an integer in [math]\displaystyle{ [n,n+\log n] }[/math], which implies in particular that the n/p_i lies within [math]\displaystyle{ O(1/\log^9 n) }[/math] of an integer. This implies that all of the 3^k signed sums

[math]\displaystyle{ \epsilon_1 n / p_1 + \ldots + \epsilon_k n / p_k }[/math]

with [math]\displaystyle{ \epsilon_i = -1,0,1 }[/math] lie within [math]\displaystyle{ O( 1 / \log^{7-o(1)} n ) }[/math] of an integer. Hopefully this sort of concentration leads to some sort of contradiction. For instance, it implies that [math]\displaystyle{ \epsilon_1/p_1 + \ldots + \epsilon_k/p_k }[/math] cannot be used to approximate too many reciprocals 1/q too closely.