It is well-known that the Mertens function $M(n)=\sum\_{k=1}^n\mu(k)$ changes sign infinitely many times when $n\rightarrow +\infty$. Let $f(n)=\sum\_{k=1}^n\frac{\mu(k)}{k}$, then $\lim\limits\_{n\rightarrow +\infty}f(n)= 0$.


**Question:** Does the function $f(n)=\sum\_{k=1}^n\frac{\mu(k)}{k}$ also change sign infinitely many times when $n\rightarrow +\infty$ ?



Sign of the function $f(n)=\sum_{k=1}^n\frac{\mu(k)}{k}$
Yes it does. To see this, note that by partial summation,
$$\frac{1}{\zeta(s + 1)} = s\int\_{1}^{\infty}\sum\_{n \leq x} \frac{\mu(n)}{n} x^{-s} \, \frac{dx}{x}$$
for all $\Re(s) > 0$. Now let $\Theta$ denote the supremum of the real part of the zeroes of $\zeta(s)$, and suppose in order to obtain a contradiction that there exists some $\varepsilon > 0$ and $x\_{\varepsilon} > 1$ such that
$$\sum\_{n \leq x} \frac{\mu(n)}{n} < x^{-1 + \Theta - \varepsilon}$$
for all $x > x\_{\varepsilon}$. Then Landau's lemma (Lemma 15.1 of Montgomery-Vaughan) states that if $\sigma\_c$ is the infimum of $\sigma \in \mathbb{R}$ for which
$$\int\_{1}^{\infty} \left(x^{-1 + \Theta - \varepsilon} - \sum\_{n \leq x} \frac{\mu(n)}{n}\right) x^{-\sigma} \, \frac{dx}{x}$$
is convergent, then
$$\int\_{1}^{\infty} \left(x^{-1 + \Theta - \varepsilon} - \sum\_{n \leq x} \frac{\mu(n)}{n}\right) x^{-s} \, \frac{dx}{x}$$
is holomorphic in the right half-plane $\Re(s) > \sigma\_c$ but not at the point $\sigma\_c \in \mathbb{R}$. On the other hand, this integral is equal to
$$\frac{1}{s + 1 - \Theta + \varepsilon} - \frac{1}{s\zeta(s + 1)}$$
for $\Re(s) > 0$ and hence for $\Re(s) > \sigma\_c$ by analytic continuation. However, this expression has a pole at $s = -1 + \Theta - \varepsilon$ and no other poles on the real line segment $\sigma > -1 + \Theta - \varepsilon$, yet by the definition of $\Theta$, there are poles in the strip $-1 + \Theta - \varepsilon < \Re(s) \leq -1 + \Theta$. Thus a contradiction is obtained, and so it follows that
$$\sum\_{n \leq x} \frac{\mu(n)}{n} = \Omega\_{+}\left(x^{-1 + \Theta - \varepsilon}\right).$$
The same method shows that
$$\sum\_{n \leq x} \frac{\mu(n)}{n} = \Omega\_{-}\left(x^{-1 + \Theta - \varepsilon}\right),$$
which implies an infinitude of sign changes. Moreover, with more work, one can show that
$$\sum\_{n \leq x} \frac{\mu(n)}{n} = \Omega\_{\pm}\left(\frac{1}{\sqrt{x}}\right),$$
and if one is sufficiently enthusiastic, then under the assumption of the Riemann hypothesis and the linear independence hypothesis,
$$\limsup\_{x \to \infty} \sqrt{x} \sum\_{n \leq x} \frac{\mu(n)}{n} = -\liminf\_{x \to \infty} \sqrt{x} \sum\_{n \leq x} \frac{\mu(n)}{n} = \infty.$$
The "true" rate of growth is probably the following:
$$0 < \limsup\_{x \to \infty} \frac{\sqrt{x}}{(\log \log \log x)^{5/4}} \sum\_{n \leq x} \frac{\mu(n)}{n} < \infty, \quad -\infty < \liminf\_{x \to \infty} \frac{\sqrt{x}}{(\log \log \log x)^{5/4}} \sum\_{n \leq x} \frac{\mu(n)}{n} < 0.$$


See also these [two](https://mathoverflow.net/questions/266391/is-the-asymptotic-for-sum-n-mun-sigman-logx-n-also-an-upper-bound/266415#266415) [answers](https://mathoverflow.net/questions/178372/lower-and-upper-bound-for-sum-k-1n-frac-1-omegakk/178375#178375) of mine.


