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
p x 1 p q
I was given that p x 1 p = log log x+O(1).
I need to show that p q x 1 p q = ( log log x ) 2 + O ( log log x )
Here we go:
Break the sum into two sums: p x 1 p q x p 1 q
Using what I was given: ( log log x + O ( 1 ) ) ( log log x p + O ( 1 ) )
Log Rules: ( log log x + O ( 1 ) ) ( log ( log x log p ) + O ( 1 ) )
Algebra: log log x log ( log x log p ) + O ( log log x )
From here I am lost. Any ideas?
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Answers (1)

Tucker House
Answered 2022-07-15 Author has 9 answers
I am assuming you mean that you are given that
(1) p x 1 p = log ( log ( x ) ) + O ( 1 )
and that you want to estimate
(2) p q x 1 p q = p x q x / p 1 p q (3) = p x 2 q x / p 1 p q (4) = p x / 2 1 p ( log ( log ( x / p ) ) + O ( 1 ) ) (5) = p x / 2 1 p log ( log ( x ) log ( p ) ) + O ( 1 ) p x 1 p (6) = p x / 2 1 p [ log ( log ( x ) ) + log ( 1 log ( p ) log ( x ) ) ] + O ( log ( log ( x ) ) ) (7) = log ( log ( x ) ) 2 + O ( log ( log ( x ) ) ) + p x / 2 1 p [ log ( 1 log ( p ) log ( x ) ) ]
Explanation:
( 2 ): write explicitly as a double sum
( 3 ): note there is no q = 1 case
( 4 ): apply ( 1 ) to the sum in q
( 5 ): incorporate the left sum into the error term from the right sum
( 6 ): apply ( 1 ) to the error term
( 7 ): x / 2 < p x 1 p log ( 2 ) and ( 1 ) imply p x / 2 1 p = log ( log ( x ) ) + O ( 1 )
It can be shown that
(8) p x / 2 1 p [ log ( 1 log ( p ) log ( x ) ) ] = O ( 1 )
but I don't see how it can be shown using just ( 1 ).

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