Riemann meets Liouville

Number Theory Level 5

$S=\sum_{n=1}^{\infty}\frac{(-1)^{\Omega(n)}}{n^4}$

Let $\Omega(n)$ denote the number of prime factors of $n$ , counted with their multiplicities. For example, $\Omega(2016)=\Omega(2^5\times 3^2\times 7)=5+2+1=8$ .

Submit $\dfrac{\pi^4}{S}$ as your answer.

The answer is 105.

1 solution

Isaac Buckley
Jan 4, 2016

$\Omega(n)$ is a (completely) additive function which implies $(-1)^{\Omega(n)}$ is a (completely) multiplicative function.

$(-1)^{\Omega(n)}$ is more commonly denoted $\lambda(n)$ and is called the Liouville function .

Since it's multiplicative we can write:

$\large S=\sum_{n=1}^{\infty} \frac{\lambda(n)}{n^4}=\prod_{p} \sum\limits_{k=0}^{\infty} \frac{\lambda(p^k)}{p^{4k}}$

Since $\lambda(p^k)=(-1)^k$ it becomes a geometric sum:

$\Large S=\prod_{p} \sum\limits_{k=0}^{\infty} \frac{(-1)^{k}}{p^{4k}}=\prod_{p} \frac{1}{1+\frac{1}{p^4}}=\prod_{p} \frac{1-\frac{1}{p^4}}{1-\frac{1}{p^8}}=\frac{\zeta(8)}{\zeta(4)}=\frac{\frac{\pi^8}{9450}}{\frac{\pi^4}{90}}=\frac{\pi^4}{105}$

Yes, another crystal clear solution, clearer than many text books! (+1)

It helps that the Liouville function is completely multiplicative: For any completely multiplicative function $f(n)$ the Euler product simplifies to $\sum_{n=1}^{\infty}\frac{f(n)}{n^s}=\prod_{p}\frac{1}{1-f(p)p^{-s}}$

Otto Bretscher - 5 years, 5 months ago

Nice identity indeed!

Pi Han Goh - 5 years, 5 months ago

Alternatively, note that $\lambda(n) = (-1)^{\Omega(n)}$ is the Liouville function, and $\displaystyle \sum_{n=1}^\infty \frac{\lambda(n)}{n^s} = \frac{\zeta(2s)}{\zeta(s)}$ which follows directly from $\lambda * 1 = \mathbf{1}_{\{ 1,4,9,16,\ldots \}}$ .

Jake Lai - 5 years, 5 months ago

Yes, this is a very elegant solution also! (+1)

Otto Bretscher - 5 years, 5 months ago

nice solution(+1). i knew this but always used dirtchlet series and never knew of this cool method!

Aareyan Manzoor - 5 years, 5 months ago

Riemann meets Liouville

The answer is 105.

1 solution

0 pending reports