# I was given that sum_(p<= x)1/p=loglogx+O(1).I need to show that sum_(pq<=x)1/(pq) = (log log x)^2+O(log log x)

Kristen Stokes 2022-07-14 Answered
$\sum _{p\le x}\frac{1}{pq}$
I was given that $\sum _{p\le x}\frac{1}{p}$ = $\mathrm{log}\mathrm{log}x$+O(1).
I need to show that $\sum _{pq\le x}\frac{1}{pq}=\left(\mathrm{log}\mathrm{log}x{\right)}^{2}+O\left(\mathrm{log}\mathrm{log}x\right)$
Here we go:
Break the sum into two sums: $\sum _{p\le x}\frac{1}{p}\sum _{q\le \frac{x}{p}}\frac{1}{q}$
Using what I was given: $\left(\mathrm{log}\mathrm{log}x+O\left(1\right)\right)\left(\mathrm{log}\mathrm{log}\frac{x}{p}+O\left(1\right)\right)$
Log Rules: $\left(\mathrm{log}\mathrm{log}x+O\left(1\right)\right)\left(\mathrm{log}\left(\mathrm{log}x-\mathrm{log}p\right)+O\left(1\right)\right)$
Algebra: $\mathrm{log}\mathrm{log}x\cdot \mathrm{log}\left(\mathrm{log}x-\mathrm{log}p\right)+O\left(\mathrm{log}\mathrm{log}x\right)$
From here I am lost. Any ideas?
You can still ask an expert for help

• Live experts 24/7
• Questions are typically answered in as fast as 30 minutes
• Personalized clear answers

Solve your problem for the price of one coffee

• Math expert for every subject
• Pay only if we can solve it

Tucker House
I am assuming you mean that you are given that
$\begin{array}{}\text{(1)}& \sum _{p\le x}\frac{1}{p}=\mathrm{log}\left(\mathrm{log}\left(x\right)\right)+O\left(1\right)\end{array}$
and that you want to estimate
$\begin{array}{}\text{(2)}& \sum _{pq\le x}\frac{1}{pq}& =\sum _{p\le x}\sum _{q\le x/p}\frac{1}{pq}\text{(3)}& & =\sum _{p\le x}\sum _{2\le q\le x/p}\frac{1}{pq}\text{(4)}& & =\sum _{p\le x/2}\frac{1}{p}\left(\mathrm{log}\left(\mathrm{log}\left(x/p\right)\right)+O\left(1\right)\right)\text{(5)}& & =\sum _{p\le x/2}\frac{1}{p}\mathrm{log}\left(\mathrm{log}\left(x\right)-\mathrm{log}\left(p\right)\right)+O\left(1\right)\sum _{p\le x}\frac{1}{p}\text{(6)}& & =\sum _{p\le x/2}\frac{1}{p}\left[\mathrm{log}\left(\mathrm{log}\left(x\right)\right)+\mathrm{log}\left(1-\frac{\mathrm{log}\left(p\right)}{\mathrm{log}\left(x\right)}\right)\right]+O\left(\mathrm{log}\left(\mathrm{log}\left(x\right)\right)\right)\text{(7)}& & =\mathrm{log}\left(\mathrm{log}\left(x\right){\right)}^{2}+O\left(\mathrm{log}\left(\mathrm{log}\left(x\right)\right)\right)+\sum _{p\le x/2}\frac{1}{p}\left[\mathrm{log}\left(1-\frac{\mathrm{log}\left(p\right)}{\mathrm{log}\left(x\right)}\right)\right]\end{array}$
Explanation:
$\left(2\right)$: write explicitly as a double sum
$\left(3\right)$: note there is no $q=1$ case
$\left(4\right)$: apply $\left(1\right)$ to the sum in $q$
$\left(5\right)$: incorporate the left sum into the error term from the right sum
$\left(6\right)$: apply $\left(1\right)$ to the error term
$\left(7\right)$: $\sum _{x/2 and $\left(1\right)$ imply $\sum _{p\le x/2}\frac{1}{p}=\mathrm{log}\left(\mathrm{log}\left(x\right)\right)+O\left(1\right)$
It can be shown that
$\begin{array}{}\text{(8)}& \sum _{p\le x/2}\frac{1}{p}\left[\mathrm{log}\left(1-\frac{\mathrm{log}\left(p\right)}{\mathrm{log}\left(x\right)}\right)\right]=O\left(1\right)\end{array}$
but I don't see how it can be shown using just $\left(1\right)$.

We have step-by-step solutions for your answer!