Analytic number theory

Lemma 1.1. If \(a_i\) is an indexed sequence of numbers \(\geq 0\), and \(f(n) = \sum _{i \leq n}a_n\), \(F(n) = \sum _{i\leq n} \log (i) a_i\). If \(F(n) >0\) for large \(n\), then \(\limsup _n{\frac {f(n)\log (n)}{F(n)}} \leq 1\). If \(f(\ceil {n^{1-\ee }}) = o(f(n))\) as \(n \to \infty \) for small \(\ee >0\), then \(\log (n)f(n) \sim F(n)\).

Proof. \(F(n) = \sum _{i \leq n}log(i)a_i \leq \sum _{i \leq n}log(n)a_n\), showing the first statement. For the second, Observe that for any \(\ee >0\), we have \(F(n) \geq \sum _{\ceil {n^{1-\ee }}}^n \log (n^{1-\ee })a_n = (1-\ee )\log (n)(f(n)-f(\ceil {n^{1-\ee }}))\), so \(\frac {F(n)}{f(n)log(n)} \geq (1-\ee )(1-\frac {f(n^{1-\ee })}{f(n)})\), and letting \(n \to \infty ,\ee \to 0\) gives the result. □