Best bound for $|\sum \prod_{p\mid n}(1-\frac{1}{p^k})-\frac m{\zeta(k+1)}|$? Follow from the previous post, we have
\begin{align*}
f_2(x):=\sum_{n\le x} \prod_{p\mid n}(1-\frac{1}{p^2})-\frac x{\zeta(3)}=O(1)\quad \text{as }x\to\infty.\tag{*}
\end{align*}
When I tried to plot $f_2$, I observed that for any $m\in \Bbb N,$
$$|f_2(m)|=\left|\sum_{n\le m}\prod_{p\mid n}(1-\frac{1}{p^2})-\frac m{\zeta(3)}\right|< c\approx0.22,$$
In particular, $0<f_2(m)<\color{red}{0.222656...}$. Here is the plot for $m\le 10000$:

I have no idea how to estimate $c\approx0.22$ through analytical method.

*

*Is it possible to find the exact value of $c$?

*If not, could we prove the less tight case (e.g. $|f_2|<0.3$)?

Since for any $\ell\ge2$,
\begin{align*}
f_\ell(x):=\sum_{n\le x} \prod_{p\mid n}(1-\frac{1}{p^\ell})-\frac x{\zeta(\ell+1)}=O(1)\quad \text{as }x\to\infty.\tag{**}
\end{align*}
One may generalize the question to finding the smallest $c_\ell$ such that
$$|f_\ell(m)|=\left|\sum_{n\le m}\prod_{p\mid n}(1-\frac{1}{p^\ell})-\frac m{\zeta(\ell+1)}\right|< c_\ell,\quad \forall m\in\Bbb N.$$


Update
Follow Gary's comment, if we define
\begin{align*}
F_2(x):=\frac1x\sum_{n\le x} f_2(n)
\end{align*}
The graph is

Mathematica code:
jordanTotient[n_, k_: 1] := 
DivisorSum[n, #^k*MoebiusMu[n/#] &] /; (n > 0) && IntegerQ[n];
Plot[1/x Sum[(Sum[jordanTotient[n, 2]/n^2, {n, 1, m}] - 
 m/Zeta[3]), {m, 1, Floor[x]}], {x, 0, 100}]


Update 2
We look at the corresponding Dirichlet series and make use of Perron's formula to get the asymptotic.
\begin{align*}
f(n)=\prod_{p\mid n}(1-\frac{1}{p^2})=\frac{J_2(n)}{n^2}\quad \implies \quad
F(s)=\sum_{n=1}^\infty\frac{f(n)}{n^s}=\sum_{n=1}^\infty\frac{J_2(n)}{n^{s+2}}=\frac{\zeta(s)}{\zeta(s+2)}.
\end{align*}
Where $J_k(\cdot)$ is the Jordan totient function. Now making use of Perron's formula,
\begin{align*}
\sum_{n=1}^m\frac{J_2(n)}{n^2}=\frac12\frac{J_2(m)}{m^2}+\frac1{2\pi i}\int_{c-i\infty}^{c+i\infty}\frac{\zeta(s)}{\zeta(s+2)}\frac{m^s}s\,\mathrm ds\tag{***}
\end{align*}
for any $c>1$. The integrand $\frac{\zeta(s)}{\zeta(s+2)}\frac{m^s}s$ has poles at $s=0$ and $s=1$.
We shift the line of integration to the left across the pole $s=1$. Let $b\in (0,1)$, we have that
\begin{align*}
\frac1{2\pi i}\int_{c-iT}^{c+iT}\frac{\zeta(s)}{\zeta(s+2)}\frac{m^s}s\,\mathrm ds=&\underbrace{\text{Res}\left(\frac{\zeta(s)}{\zeta(s+2)}\frac{m^s}s,1\right)}_{=m/\zeta(3)}+
\underbrace{\frac1{2\pi i}\int_{b-iT}^{b+iT}\frac{\zeta(s)}{\zeta(s+2)}\frac{m^s}s\,\mathrm ds}_{\text{vertical integral}}\\
&+ \underbrace{\frac1{2\pi i}\int_{c-iT}^{b-iT}\frac{\zeta(s)}{\zeta(s+2)}\frac{m^s}s\,\mathrm ds+\frac1{2\pi i}\int_{b+iT}^{c+iT}\frac{\zeta(s)}{\zeta(s+2)}\frac{m^s}s\,\mathrm ds}_{\text{horizontal integral}}.
\end{align*}
Thus,
$$f_2(m)=\sum_{n=1}^m\frac{J_2(n)}{n^2}-\frac{m}{\zeta(3)}=\frac12\frac{J_2(m)}{m^2}+\color{red}{O(?)}$$
but I do not know how to estimate these integrals to get $O(?)$.
I am not sure if the above steps are correct as I am new on this topic. Any help would be much appreciated!
 A: For convenience, let
$$
g_\ell(n)=\prod_{p|n}\left(1-{1\over p^\ell}\right),
$$
so that it follows from the Dirichlet convolution property
$$
g_\ell(n)=\sum_{d|n}{\mu(d)\over d^\ell}
$$
that
\begin{aligned}
\sum_{n\le x}g_\ell(n)
&=\sum_{d\le x}{\mu(d)\over d^\ell}\left\lfloor\frac xd\right\rfloor=x\sum_{d\le x}{\mu(d)\over d^{\ell+1}}-\sum_{d\le x}{\mu(d)\over d^\ell}\left\{\frac xd\right\} \\
&={x\over\zeta(\ell+1)}-x\sum_{d>x}{\mu(d)\over d^\ell}-\sum_{d\le x}{\mu(d)\over d^\ell}\left\{\frac xd\right\}.
\end{aligned}
This suggests that the key ingredients in the estimation of $f_\ell(m)$ are as follows:
$$
|f_\ell(x)|\le x|S_1(x)|+|S_2(x)|.
$$
where
$$
S_1(x)=\sum_{d>x}{\mu(d)\over d^\ell},\quad S_2(x)=\sum_{n\le x}{\mu(d)\over d^\ell}\left\{\frac xd\right\}.
$$
Estimation of $S_1(x)$
The prime number theorem in the form of de la Vallée Poussin implies that there exists an constant $c_0>0$ such that
$$
M(x)=\sum_{n\le x}\mu(n)=O\left\{x\exp\left(-c_0\sqrt{\log x}\right)\right\}
$$
This indicates that
$$
\sum_{d>x}{\mu(d)\over d^\ell}=\int_x^\infty{\mathrm dM(t)\over t^\ell}=-{M(x)\over x^\ell}+\ell\int_x^\infty{M(t)\over t^{\ell+1}}\mathrm dt=O\left\{x^{1-\ell}\exp\left(-c_0\sqrt{\log x}\right)\right\}
$$
If $\ell\ge2$ then $x|S_1(x)|$ will certainly vanish at infinity. This suggests that for all $\varepsilon>0$ there exists an $x_0=x_0(\varepsilon)$ such that for all $x>x_0$ there is
$$
|f_\ell(x)|<|S_2(x)|+\varepsilon
$$
Thus, the remaining task is to handle $S_2(x)$.
Preliminary treatments for $S_2(x)$
\begin{aligned}
S_2(x)
&=\frac12\sum_{d\le x}{\mu(d)\over d^\ell}+S_3(x) \\
&={1\over2\zeta(\ell)}+O\left\{x^{1-\ell}\exp\left(-c_0\sqrt{\log x}\right)\right\}+S_3(x)
\end{aligned}
where
$$
S_3(x)=\sum_{d\le x}{1\over d^\ell}\overline B_1\left(\frac xd\right)
$$
and $\overline B_n(x)$ denotes the periodized Bernoulli polynomials. I guess tricks from the theory of trigonometric sums/integral should come in handy to handle $S_3(x)$.
