That's a weird sum

Number Theory Level 5

$\sum _{ n=1 }^{ \infty } \frac { { 0.5 }^{ \Omega (n) } }{ { n }^{ 2 } } \sum _{ k=1 }^{ \infty } \frac { \mu \left( k \right) { 0.5 }^{ \Omega (k) } }{ { k }^{ 2 } }$

$\mu(n)$ is the mobius function and $\Omega (n)$ outputs the number of prime divisors of $n$ counted with multiplicity.

$\textbf{Bonus:}$ Find a closed form for the sum, (which I am unable to do)

$\sum _{ n=1 }^{ \infty } \frac { { 2 }^{ \Omega (n) } }{ { n }^{ 2 } }$

The answer is 1.00.

1 solution

Otto Bretscher
Jan 14, 2016

The product of these two Dirichlet series is the Dirichlet series of the convolution of the numerators. If we let $f(n)=0.5^{\Omega(n)}$ , a totally multiplicative function, then $f*(f\mu)=f(1*\mu)=f\epsilon$ where $\epsilon(1)=1$ and $\epsilon(n)=0$ for $n>1$ . Thus the product we seek is $\sum_{n=1}^{\infty}\frac{f(n)\epsilon(n)}{n^2}=\boxed{1}$ .

Bonus: $\sum_{n=1}^{\infty}\frac{2^{\Omega(n)}}{n^2}=\prod_{p}\frac{1}{1-\frac{2}{p^2}}=\frac{1}{2C-1}$ where $C$ is the Feller Tornier constant .

Moderator note:

Great explanation. Recognizing how Dirichlet functions interact is important to understanding their theory.

@Julian Poon , you told us that the prime product was yet to be found!!

Aareyan Manzoor - 5 years, 5 months ago

I said that the prime product was already known as the Feller Tornier constant, but I can't find a closed form for it.

Julian Poon - 5 years, 5 months ago

@Julian Poon Where did you say that the product is known as the Feller-Tornier constant? I must have missed that ;)

Otto Bretscher - 5 years, 5 months ago

@Otto Bretscher – A solution was deleted and it is in the comments section I said that.