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

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- Solution d

Solution

Taking the natural logarithm of the finite Euler product and using the Taylor series

- lo g (1 - u) = u + \sum_{k \geq 2} \frac{u ^{k}}{k}

gives

lo g \prod_{p \leq x} (1 - \frac{1}{p})^{- 1} = \sum_{p \leq x} \frac{1}{p} + \sum_{p \leq x} \sum_{k \geq 2} \frac{1}{k p ^{k}} .

The double series converges absolutely, and the Mertens second theorem therefore makes the right side

lo g lo g x + lo g C + O (\frac{1}{l o g x})

for

C = exp (c + \sum_{p} \sum_{k \geq 2} \frac{1}{k p ^{k}}) > 0.

Exponentiating proves

p \leq x \prod (1 - \frac{1}{p})^{- 1} = C lo g x + O (1) .

This is the Mertens third theorem.

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