# Math Genius: Asking for help in solving a summation in number theory

Summation is S = $$sum_{q, l} 1$$ subject to conditions (11) and (13) which could be written as $$sum_{q,l , (11), (13) } 1$$, where (11) is $$q leq log^{3n} n$$ and (13) is $$-1/2< x-l/q leq 1/2.$$ Here $$n$$ is an integer sufficiently large, $$l$$ belongs to integers and $$q$$ belongs to positive integers, $$x in mathbb R,$$ and $$gcd$$(l, q) =1.\$
It is to be proved that $$sum_{q,l , (11), (13) } 1$$ = $$sum_{q, (11) } q < log^{6n} n$$ .
$$S = sum limits_{q,l, (q,l)=1} 1 leq sum limits_{q,l} 1$$ and we have that $$-0.5 leq x-frac{l}{q} leq 0.5$$ so $$x-0.5leq frac{l}{q} leq x+0.5$$ so $$q(x-0.5) leq l leq q(x+0.5)$$ and we have that $$S = sum limits_{q} sumlimits_{l = q(x-0.5)}^{q(x+0.5)}1$$ and we have $$q(x+0.5)-q(x-0.5)$$ many times ‘1’ so we have that $$sumlimits_{l = q(x-0.5)}^{q(x+0.5)}1 = q$$ and then we have that $$sum limits_{qleq ln^{3n} n} q = sum limits_{q=1}^{ln^{3n}n} q = frac{ln^{3n}n(ln^{3n}n+1)}{2} leq ln^{6n}n$$ for all $$n geq 3$$.
Note : the above is valid becuase we are looking for upper bound and the fact that without the condition that $$(q,l)=1$$ we make the sum $$S$$ bigger.