Home » Uncategorized » On a strange Möbius series

On a strange Möbius series

Let \mu denote the Möbius function. Evaluate the series

    \[\mathcal{S} = \sum_{n=1}^\infty\frac{(-1)^{\mu(n)}}{n^s}\]

where \mathfrak{Re}(s)>1.

Solution

Since

    \[(-1)^{\mu(n)}= 1 - 2\mu^2(n)\]

we deduce that

\displaystyle \sum_{n= 1}^{\infty} \frac{(-1)^{\mu(n)}}{n^s} = \zeta(s)-2\sum_{n\geq 1}\frac{\mu^2(n)}{n^s} = \zeta(s)-\frac{2\,\zeta(s)}{\zeta(2s)}

Read more

Leave a comment

Who is Tolaso?

Find out more at his Encyclopedia Page.

Donate to Tolaso Network