This appears in this question: Show that $\sum_{pq\leq x}\frac{1}{pq}$ = $(\ln \ln x)^2+O(\ln \ln x)$
Here, $p$ is a prime that is less than a natural number $x$.
In that question, it is said that $\log \left(1 – \frac{\log p}{\log {x}} \right) = O\left(\frac{\log p}{\log x}\right)$.
However, I am not aware of a way to do it.
If the above estimation is too tight, then I only wish to bound it to be
$$\log \left(1 – \frac{\log p}{\log {x}} \right) = o(\log \log x)$$
Any suggestions?
Best Answer
We have
$$\log(1-y)=-y-\frac{y^2}{2}-\frac{y^3}{3}-\ldots,$$ by Maclaurin series, where $y<1$. Then
$$\log(1-y)=-y+O(y^2)=O(y),$$ where we have $|y| < 1$, as higher order terms go to zero quickly.
Since $p < x$ and $\log$ is an increasing function, we have $\log(p)<\log(x)$ or $0 < \frac{\log(p)}{\log(x)} < 1$, assuming $x \gt p \gg 1$. Consequently, $$ \log\left(1-\frac{\log(p)}{\log(x)}\right)=O\!\left(\frac{\log(p)}{\log(x)}\right).$$