Best bound for $|\sum \prod_{p\mid n}(1-\frac{1}{p^k})-\frac m{\zeta(k+1)}|$?

Question

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 $$|f_\ell(m)|=\left|\sum_{n\le m}\prod_{p\mid n}(1-\frac{1}{p^\ell})-\frac m{\zeta(\ell+1)}\right|< c_\ell,\quad \forall m\in\Bbb N.$$

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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}]

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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!

Answer 1

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 $S_3(x)$.