Mertens' function

Question

I am tasked with applying the Wiener-Ikehara Theorem to achieve a bound of little o(x) on Mertens' function $\sum_{n=1}^x \mu (n)$. My problem is the Wiener-Ikehara Theorem applies to Dirichlet series $\sum_{n=1}^\infty a(n) n^{-s}$ only when $a(n)$ is non-negative but clearly $\mu(n)$ can be negative although its absolute value is bounded by $1$. I also know this Dirichlet series can b e extended to a meromorphic function in the region $\mathbf{R}(s) \geq1$ having only a simple pole at s=1 with residue $R\geq0$ which gives $\sum_{n=1}^x a(n) =Rx+ o(x)$ as x goes to $\infty$. But I am unsure as to how I should work around the fact that my a(n) will not always be non negative.

Answer 1

The Möbius function isn't non-negative, but it is bounded. Hence we obtain a non-negative sequence by adding a suitable constant to it. With $a(n) = 1 + \mu(n)$, we have

$$\sum_{n = 1}^{\infty} \frac{a(n)}{n^s} = \zeta(s) + \frac{1}{\zeta(s)}$$

for $\operatorname{Re} s > 1$. Since $a(n) \geqslant 0$ and $\zeta(s)$ has no zeros on the line $\operatorname{Re} s = 1$, the assumptions of the Wiener-Ikehara theorem are satisfied, whence

$$\sum_{n \leqslant x} a(n) \sim x.$$

Consequently

$$\frac{M(x)}{x} = \frac{\sum_{n\leqslant x} a(n) - \lfloor x\rfloor}{x} \xrightarrow{x\to +\infty} 1 - 1 = 0.$$