Check the solution to "Proof that Dirichlet series∑n=1∞2w(n)n2=52"

Ingrid Senior

Answered question

2022-02-25

Proof that Dirichlet series

\sum_{n = 1}^{\infty} \frac{2^{w (n)}}{n^{2}} = \frac{5}{2}

Clay Giles

Beginner2022-02-26Added 4 answers

2^{w} = 1 \cdot μ^{2}

therefore

\sum_{n = 1}^{+ \infty} \frac{2^{w (n)}}{n^{2}} = \sum_{n = 1}^{+ \infty} \frac{1}{n^{2}} \sum_{n = 1}^{+ \infty} \frac{μ {(n)}^{2}}{n^{2}}

= ζ (2) \frac{ζ (2)}{ζ (4)}

= \frac{π^{2}}{6} \frac{\frac{π^{2}}{6}}{\frac{π^{4}}{90}}

= \frac{5}{2}

Indeed,

ζ (2 s) \sum_{n = 1}^{+ \infty} \frac{μ {(n)}^{2}}{n^{s}} = \sum_{n, m = 1}^{+ \infty} \frac{μ {(n)}^{2}}{{(n m^{2})}^{s}}

= \sum_{p = 1}^{+ \infty} \frac{1}{p^{s}} = ζ (s)

because every integer

p \geq 1

can be uniquely written as

p = n m^{2}

where

m \geq 1

and

n \geq 1

is a squarefree integer.

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