The book "The Riemann Hypothesis, A Resource for the Afficionado and Virtuoso Alike", Borwein, Choi, Rooney, Weirathmueller, Eds., states on its page six the following theorem (Theorem 1.2):

The Riemann hypothesis is equivalent to the statement that for every fixed $\epsilon > 0$,

$$\lim_{n \to \infty} \frac{\lambda(1) + \lambda(2)+ ...+\lambda(n)}{n^{\frac 12 + \epsilon}} = 0.$$

(Here, $\lambda$ is the Liouville function $\lambda: n \mapsto (-1)^{\omega(n)}$ where $\omega(n)$ is the number of, not necessarily distinct, prime factors of $n$.) The editors remark that this statement among others were considered by Landau in his doctoral thesis of 1899. There is translation by Coons of a dissertation of Landau, which is described as his doctoral dissertation, here:


But I haven't been able to discern a direct connection between this article and the theorem above. Am I missing something? Is there perhaps a proof of this theorem available somewhere else? I've looked a bit, and I haven't found one.

Reason for asking: I wonder, under the Riemann hypothesis, what happens to the exponent in the denominator if powers of $-1$ in the definition of the Liouville function are replaced by powers of some other root of unity $\exp(2 \pi i/k), k >2$.

  • 1
    $\begingroup$ Work out why the Dirichlet series for the multiplicative function $\lambda(n)$ is $\zeta(2s)/\zeta(s)$ and see how those calculations change if you use the multiplicative function $\lambda_z(n)$ instead where on prime powers $\lambda_z(p^r) = z^r$ for a root of unity $z$ with order $k$. How does the identity $1/(1+x) = (1-x)/(1-x^2)$ change when the left side is $1/(1-zx)$? $\endgroup$
    – KConrad
    May 11, 2017 at 2:34

1 Answer 1


About your reason for asking. See https://doi.org/10.1215/00277630-1260423, https://projecteuclid.org/euclid.nmj/1306851587 or the arxiv version https://arxiv.org/pdf/0906.1029.pdf where such things are discussed. In particular, the partial sums $\sum_{n\leq x}\exp(2\pi i\Omega(n)/k)$ are not $o(x^\alpha)$ for any $\alpha<1$.


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