Consider the function $a(n)$ defined in formula (1) below, the related summatory functions defined in formulas (2) to (5) below, and their relationships with the Riemann zeta function $\zeta(s)$ defined in formulas (6) to (9) below.
$$a(n)=\sum\limits_{d|n}\mu(d)\ \mu\left(\frac{n}{d}\right)\tag{1}$$
$$f(x)=\sum\limits_{n=1}^x a(n)\tag{2}$$
$$g(x)=\sum\limits_{n=1}^x\frac{a(n)}{n}\tag{3}$$
$$M(x)=\sum\limits_{n=1}^x\mu(n)=\sum\limits_{n=1}^x a(n)\left\lfloor\frac{x}{n}\right\rfloor\tag{4}$$
$$h(x)=\sum\limits_{n=1}^x\frac{\mu(n)}{n}\tag{5}$$
$$s\int_0^\infty f(x)\ x^{-s-1}\ dx=\underset{N\to\infty}{\text{lim}}\left(\sum_{n=1}^N a(n)\ n^{-s}\right)=\frac{1}{\zeta(s)^2}\ ,\quad\Re(s)>1\tag{6}$$
$$s\int_0^\infty g(x)\ x^{-s-1}\ dx=\underset{N\to\infty}{\text{lim}}\left(\sum_{n=1}^N a(n)\ n^{-s-1}\right)=\frac{1}{\zeta(s+1)^2}\ ,\quad\Re(s)>0\tag{7}$$
$$s\int_0^\infty M(x)\ x^{-s-1}\ dx=\underset{N\to\infty}{\text{lim}}\left(\sum_{n=1}^N a(n)\ n^{-s-1}\right)=\frac{1}{\zeta(s)}\ ,\quad\Re(s)>1\tag{8}$$
$$s\int_0^\infty h(x)\ x^{-s-1}\ dx=\underset{N\to\infty}{\text{lim}}\left(\sum_{n=1}^N a(n)\ n^{-s-1}\right)=\frac{1}{\zeta(s+1)}\ ,\quad\Re(s)>0\tag{9}$$
I know the functions $f(x)$ and $M(x)$ are unbounded and I know $\underset{x\to\infty}{\text{lim}}\ h(x)=0$, but I'm wondering if $\underset{x\to\infty}{\text{lim}}\ g(x)=0$.
Question (1): Is it true $\underset{x\to\infty}{\text{lim}}\ g(x)=0$?
If the answer to question (1) is no, I'm wondering if the function $g(x)$ can be evaluated arbitrarily close to zero as $x\to\infty$. Here is a list of values of $x$ and $g(x)$ starting at $x=3$ where the magnitude of $g(x)$ in each subsequent entry in the list is smaller than the magnitudes of $g(x)$ for all smaller values of $x\ge 3$. The list was started at $x=3$ because $g(2)=0$ which I suspect is the only zero of $g(x)$. The values of $g(x)$ were search all the way up to $x=1000000$.
$$\begin{array}{cc}
x & g(x) \\
3 & -0.666667 \\
4 & -0.416667 \\
6 & -0.15 \\
10 & 0.0753968 \\
17 & 0.00779972 \\
42 & -0.00552823 \\
106 & 0.00042345 \\
516 & 0.000249329 \\
1230 & 0.000111979 \\
1234 & 0.000106715 \\
1260 & -0.0000321944 \\
1267 & -\text{9.63619537368769$\grave{ }$*${}^{\wedge}$-6} \\
28673 & \text{1.0486208171857959$\grave{ }$*${}^{\wedge}$-6} \\
163635 & -\text{1.0115813722810309$\grave{ }$*${}^{\wedge}$-6} \\
213491 & -\text{8.539627303654234$\grave{ }$*${}^{\wedge}$-7} \\
213873 & -\text{6.915942594565252$\grave{ }$*${}^{\wedge}$-7} \\
213877 & -\text{6.912444779115509$\grave{ }$*${}^{\wedge}$-7} \\
213966 & -\text{6.274832151405875$\grave{ }$*${}^{\wedge}$-7} \\
214358 & -\text{1.3308289434561706$\grave{ }$*${}^{\wedge}$-7} \\
214362 & -\text{1.329958423818724$\grave{ }$*${}^{\wedge}$-7} \\
500339 & \text{8.99566874867865$\grave{ }$*${}^{\wedge}$-8} \\
513326 & \text{1.9952399234664665$\grave{ }$*${}^{\wedge}$-8} \\
552079 & \text{1.022984743834322$\grave{ }$*${}^{\wedge}$-8} \\
552083 & \text{1.0190476629886284$\grave{ }$*${}^{\wedge}$-8} \\
552162 & \text{4.955322537652748$\grave{ }$*${}^{\wedge}$-9} \\
\end{array}$$
Question (2): If the answer to question (1) is no, can the function $g(x)$ can be evaluated arbitrarily close to zero as $x\to\infty$?
Best Answer
It follows from the prime number theorem that $\sum_{n\ge 1} \frac{\mu(n)}n$ and related series such as $\sum_{n\ge 1} \frac{\sum_{d|n} \mu(d) \mu(n/d)}n$ converge.
The convergence of $\sum_{n\ge 1} \frac{\mu(n)}n$ implies the PNT, so does the convergence of $\sum_{n\ge 1} \frac{\sum_{d|n} \mu(d) \mu(n/d)}n$ (can you show it?).
It is elementary (partial summation) that if $\sum_{n\ge 1} \frac{b_n}n$ converges then the limit is $\lim_{s\to 1^+}\sum_{n\ge 1} \frac{b_n}{n^s}$, whence your series converges to $0$.