Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 150 / 1 / c

Table of contents
- Solution c

Solution

The same divisor-identity argument as in part (a) gives

\sum_{n \leq x} \frac{Λ ( n )}{n} = lo g x + O (1) .

On the other hand, the Abel summation formula gives

\sum_{n \leq x} \frac{Λ ( n )}{n} = \frac{ψ ( x )}{x} + \int_{1}^{x} \frac{ψ ( t )}{t ^{2}} d t .

Under the proposed asymptotic, the right side is

a lo g x + b lo g lo g x + o (lo g lo g x) + O (1) .

Dividing first by

lo g x

gives

a = 1

. Subtracting

lo g x

, dividing by

lo g lo g x

, and taking the limit then gives

b = 0

. Therefore

a = 1, b = 0.

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