On the Means of Jordan’s Totient Function

Abstract

In this paper, we study some important means of Jordan’s totient function, especially, we obtain asymptotic formula for geometric mean and harmonic mean. We also study alternating sums of Jordan’s totient function and Carleman’s inequality for this function.

1 Introduction and Summary of the Results

Euler’s totient function $\varphi$ is one of the most famous arithmetic functions used in number theory. Recall that $\varphi (n)$ is defined as the number of positive integers less than or equal to n that are coprime to n. Many generalizations and analogs of Euler’s function are known. See the special chapter on this topic in [15] and the references given there. Among the generalizations, the most significant is probably Jordan’s totient function. For given positive integer r, Jordan’s function $J_r$ is defined by the number of r-tuples of integers $(a_1, a_2, \dots , a_r)$ satisfying $1\leqslant a_i \leqslant n$ and $\gcd (a_1, a_2, \dots , a_r, n)=1$. It can be shown that (by inclusion–exclusion principle)

$$\begin{aligned} J_r(n)= n^r\prod _{p|n}\left( 1-\frac{1}{p^r}\right) . \end{aligned}$$

(1.1)

The function $J_r$ is multiplicative and if p is a prime and $\alpha$ is a positive integer, then $J_r(p^{\alpha })= p^{r\alpha }- p^{r(\alpha -1)}$. Dirichlet series related to Jordan’s function is as follows:

$$\begin{aligned} \sum _{n=1}^{\infty }\frac{J_r(n)}{n^s}=\frac{\zeta (s-r)}{\zeta (s)}\quad (\mathfrak {R}(s)> r+1), \end{aligned}$$

where $\zeta (s)$ is the Riemann zeta-function, defined by $\zeta (s)=\sum _{n=1}^{\infty }\frac{1}{n^s}$ for $\mathfrak {R}(s)>1$.

Gauss type formula for $J_r$ reads as follows:

$$\begin{aligned} \sum _{d|n}J_r(d)=n^r, \end{aligned}$$

from which, the Möbius inversion formula gives

$$\begin{aligned} J_r(n)=\sum _{d|n}\mu (d)\left( \frac{n}{d}\right) ^r. \end{aligned}$$

(1.2)

For more results and applications of this function, as well as its history, we refer the reader to [2, 3, 13, 16, 19].

In this note, we study the means of Jordan’s totient function. The asymptotic expansion of arithmetic mean of $J_r$ is known in the literature (for example, see [13]) and it is shown that for $r>1$,

$$\begin{aligned} A_{n}(J_r):=\frac{1}{n} \sum _{k\leqslant n}J_r(k)= \frac{n^{r }}{(r+1)\zeta (r+1)} + O\Big ( n^{r-1} \Big ). \end{aligned}$$

(1.3)

In this note, we obtain the asymptotic behaviour of geometric mean of the values of Jordan’s function. More precisely, we prove the following.

Theorem 1.1

Assume that $r\geqslant 1$ is a fixed integer. Then

$$\begin{aligned} \sum _{k\leqslant n}\log J_r(k)= r n\log n +(\alpha _r -r)n +\frac{r}{2}\log n + R(n), \end{aligned}$$

(1.4)

where

$$\begin{aligned} \alpha _r = \sum _{p }\frac{1}{p}\log \Big (1-\frac{1}{p^r}\Big ), \end{aligned}$$

and

$$\begin{aligned} R(n)= {\left\{ \begin{array}{ll} O(\log \log n) &{} \text {if}\quad r=1, \\ O_r(1) &{} \text {if} \quad r>1 . \end{array}\right. } \end{aligned}$$

Corollary 1.2

For $r\geqslant 1$, we have

$$\begin{aligned} G_{n}(J_r)= B_rn^r + \frac{r}{2}B_r n^{r-1}\log n + E(n), \end{aligned}$$

(1.5)

where

$$\begin{aligned} B_r=e^{\alpha _r -r}= \frac{1}{e^r}\prod _{p}\left( 1-\frac{1}{p^r}\right) ^{\frac{1}{p}}, \end{aligned}$$

and

$$\begin{aligned} E(n)= {\left\{ \begin{array}{ll} O(\log \log n) &{} \text {if}\quad r=1, \\ O_r( n^{r-1}), &{} \text {if}\quad r>1 . \end{array}\right. } \end{aligned}$$

We mention that the case $r=1$ in the above corollary, which corresponds Euler’s function, has been obtained already in [10].Regarding the harmonic mean, we recall that [12],

$$\begin{aligned} \sum _{n\leqslant x}\frac{1}{\varphi (n)}=\frac{\zeta (2)\zeta (3)}{\zeta (6)}\left( \log x +\gamma -\sum _{p}\frac{\log p}{p^2+p-1}\right) + O\Big ( \frac{\log x}{x} \Big ), \end{aligned}$$

where $\gamma$ is Euler’s constant.

For more results about reciprocal sums of arithmetic functions, we refer the reader to [4, 7]. For the reciprocal sum of Jordan’s function, we prove the following.

Theorem 1.3

Assume that $r>1$ is a fixed integer. As $x \rightarrow \infty$, we have

$$\begin{aligned} \sum _{n\leqslant x}\frac{1}{J_r(n)}=\zeta (r) \beta _r -\frac{\beta '_r}{(r-1)x^{r-1}}+ O\Big (\frac{1}{x^r}\Big ), \end{aligned}$$

where

$$\begin{aligned} \beta _r:= \prod _{p}\left( 1+\frac{1}{p^r(p^r-1)}\right) \quad \text {and} \quad \beta '_r:=\prod _{p}\left( 1+\frac{1}{p(p^r-1)}\right) . \end{aligned}$$

Moreover, reciprocal sum of Euler $\varphi$ function running over squarefree integers has been studied. (For example, see [17].)

$$\begin{aligned} \sum _{n\leqslant x}\frac{\mu ^2(n)}{\varphi (n)}=\log x + \gamma +\sum _{p}\frac{\log p}{p(p-1)}+O\Big (\frac{1}{\sqrt{x}}\Big ). \end{aligned}$$

We obtain asymptotic formula for reciprocal sum of Jordan’s function running over squarefree integers.

Theorem 1.4

For $r>1$, we have

$$\begin{aligned} \sum _{n\leqslant x}\frac{\mu ^2(n)}{J_r(n)}= \frac{r\gamma _r}{r-1}-\frac{\gamma '_rx^{1-r}}{r-1}+O\left( x^{1/2-r}\right) , \end{aligned}$$

where

$$\begin{aligned} \gamma _r = \frac{1}{\zeta (2)}\prod _{p}\left( 1+\frac{1}{p^{r-1}(p^r-1)(p+1)}\right) \quad \text {and} \quad \gamma '_r =\prod _p\left( 1-\frac{p^r-p}{p^2(p^r-1)}\right) . \end{aligned}$$

Recently in [18], L.Tóth studied alternating sums of multiplicative arithmetic functions; similarly, we study alternating sums concerning Jordan’s function. We state the following theorems.

Theorem 1.5

Assume that $r\geqslant 1$ is a fixed integer, we have

$$\begin{aligned} \sum _{n\leqslant x}(-1)^{n-1}J_r(n)=\frac{1}{(2^{r+1}-1)(r+1)\zeta (r+1)}x^{r+1}+O\Big ( x^r \Big ). \end{aligned}$$

Theorem 1.6

Assume that $r >1$ is a fixed integer, we have

$$\begin{aligned} \sum _{n\leqslant x}(-1)^{n-1}\frac{1}{J_r(n)}= \zeta (r) \beta _r \beta ''_r + \frac{\beta '_r}{(1-2^{r+1})(1-r)x^{r-1}}+O\Big (\frac{1}{x^r}\Big ), \end{aligned}$$

where $\beta _r$ and $\beta '_r$ are defined as in Theorem 1.3, and

$$\begin{aligned} \beta ''_r=\frac{2}{\sum _{\nu =0}^{\infty }\frac{1}{J_r(2^{\nu })}}-1= 2\left( 1-2^{-r}\right) ^2 -1. \end{aligned}$$

For positive real numbers $a_1, a_2, \dots ,a_n$, Carleman’s inequality [5] asserts that

$$\begin{aligned} \sum _{k=1}^{n}(a_1\dots a_k)^{\frac{1}{k}}\leqslant e\,\sum _{k=1}^{n}a_k. \end{aligned}$$

(1.6)

The constant e is the best possible. Recently in [11], Hassani studied Carleman’s inequality over the values of arithmetical functions. Hassani studied Carleman’s inequality over prime numbers and over reciprocal of the prime numbers, and showed that

$$\begin{aligned}&\frac{\sum _{k=1}^{n}\left( p_1 \dots p_k\right) ^{\frac{1}{k}}}{\sum _{k=1}^{n}p_k}=\frac{1}{e} +O\left( \frac{1}{\log n}\right) \quad \textit{and}\\&\quad \times \frac{\sum _{k=1}^{n}\left( \frac{1}{p_1}\dots \frac{1}{p_k}\right) ^{\frac{1}{k}}}{\sum _{k=1}^{n}\frac{1}{p_k}}=e +O\left( \frac{1}{\log \log n}\right) . \end{aligned}$$

For more results, we refer the reader to [11]. For each positive arithmetical function f, let

$$\begin{aligned} C_f(n):= \frac{\sum _{k=1}^{n}(f(1)\dots f(k))^{\frac{1}{k}}}{\sum _{k=1}^{n}f(k)}. \end{aligned}$$

We can write

$$\begin{aligned} C_f(n)=\frac{\sum _{k=1}^{n}G_k(f)}{\sum _{k=1}^{n}f(k)}. \end{aligned}$$

(1.7)

We prove the following results:

Theorem 1.7

Assume that $r >1$ is a fixed integer, as $n\rightarrow \infty$ we have

$$\begin{aligned} C_{J_r}(n)= B_r\zeta (1+r)+\frac{B_r(r+1)}{2}\frac{\log n}{n}+O\left( \frac{1}{n}\right) , \end{aligned}$$

where $B_r$ is absolute constant defined in Corollary 1.2.

Remark 1.8

The asymptotic behaviour of the ratio of the arithmetic to the geometric means of several numbers of theoretic sequences is studied by M.Hassani (for example, see [8]). As a result, for $r>1$, as $n \rightarrow \infty$, we have

$$\begin{aligned} \frac{A_{n}(J_r)}{G_{n}(J_r)}=\frac{1}{B_r\,(r+1)\zeta (r+1)}\left( 1+ \frac{r}{2}\frac{ \log n}{n}+O\Big (\frac{1}{ n}\Big )\right) . \end{aligned}$$

2 Preliminaries

To prove our results, we need some results and notations.

Stirling’s theorem asserts that (for more details, see [9])

Lemma 2.1

As n tends to infinity, we have

$$\begin{aligned} \sum _{k=1}^{n}\log k = n\log n -n +\frac{1}{2}\log n + C + O\left( \frac{1}{n}\right) , \end{aligned}$$

where

$$\begin{aligned} C= \log \sqrt{2\pi }. \end{aligned}$$

Theorem 3.2 of [1] asserts that

Lemma 2.2

For $\alpha >0$ and $\alpha \ne 1$, we have

$$\begin{aligned} \sum _{n\leqslant x}\frac{1}{n^{\alpha }}= \frac{x^{1-\alpha }}{1-\alpha }+ \zeta (\alpha ) + O\Big (\frac{1}{x^{\alpha }}\Big ). \end{aligned}$$

Lemma 1.3 of [8] asserts that

Lemma 2.3

for each real $\alpha > 1$ and each real $z > 1$, we have

$$\begin{aligned} \sum _{p>z}\frac{1}{p^{\alpha }}\ll \frac{1}{z^{\alpha -1}\log z}. \end{aligned}$$

Chapter 1, Section 1.5 of [14] asserts that

Lemma 2.4

For $r \geqslant 2$, there are positive constants $c_1$ and $c_2$ such that

$$\begin{aligned} c_1 n^r \leqslant J_r(n) \leqslant c_2 n^r. \end{aligned}$$

In fact, one can show that $c_1=\frac{1}{\zeta (r)}$ and $c_2=1$.

Lemma 2.5

For $r \geqslant 1$, we have

$$\begin{aligned} \frac{n^r}{J_r(n)}=\sum _{d|n}\frac{\mu ^2(d)}{J_r(d)}. \end{aligned}$$

(2.1)

Lemma 5.2 of [6] asserts that

Lemma 2.6

For each positive integer k, we have

$$\begin{aligned} \sum _{\begin{array}{c} n\leqslant x\\ (n, k)=1 \end{array}}\mu ^2(n)=\frac{xk}{\zeta (2)\psi (k)}+O\Big (\theta (k)\sqrt{x}\Big ), \end{aligned}$$

where $\psi$ is Dedekind’s psi function, defined by $\psi (n)=n\prod _{p|n}\Big (1+\frac{1}{p}\Big )$, and $\theta (n)=2^{\omega (n)}$, where $\omega (n)$ denotes the number of distinct prime divisors of n.

Assume that f is a nonzero complex-valued multiplicative function. Let

$$\begin{aligned} D(f,s):=\sum _{n=1}^{\infty }\frac{f(n)}{n^s} \end{aligned}$$

denote the Dirichlet series of the function f. Then (see [18, Proposition 1])

$$\begin{aligned} \sum _{n=1}^{\infty }(-1)^{n-1}\frac{f(n)}{n^s}=D(f,s)\left( 2\left( \sum _{\nu =0}^{\infty }\frac{f(2^{\nu })}{2^{\nu s}}\right) ^{-1}-1\right) . \end{aligned}$$

(2.2)

Analogous notations and methods in [18], consider the formal power series

$$\begin{aligned} S_f(x):=\sum _{\nu =0}^{\infty }a_{\nu }x^{\nu }, \end{aligned}$$

where $a_{\nu }=f(2^{\nu })\,\,(\nu \geqslant 0)$, $a_0 = f(1) = 1$. Note that $S_f(x)$ is the Bell series of the function f for the prime $p = 2$. Let

$$\begin{aligned} \overline{S}_f(n):=\sum _{\nu =0}^{\infty }b_{\nu }x^{\nu } \end{aligned}$$

be its formal reciprocal power series. Here the coefficients $b_{\nu }$ are given by $b_0= 1$ and $\sum _{j=0}^{\nu }a_jb_{\nu -j}=0\,\, (\nu \geqslant 0)$.

It follows from (2.2) that the convolution identity

$$\begin{aligned} (-1)^{n-1}f(n)=\sum _{d|n}h_f(d)f\left( \frac{n}{d}\right) \quad (n\geqslant 1), \end{aligned}$$

(2.3)

holds, where the function $h_f$ is multiplicative, $h_f(p^{\nu })=0$ if $p > 2, \nu \geqslant 1$ and $h_f(2^{\nu })=2b_v \, (\nu \geqslant 1), \,\, h_f(1)=2b_0-1=1$. Now, we apply (2.2) and (2.3) to get

$$\begin{aligned} \sum _{n=1}^{\infty }\frac{h_f(n)}{n^s}=\frac{2}{S_f(1/2^s)}-1. \end{aligned}$$

Thus, for each $r\geqslant 1$, and for any real number $s>r$, we obtain

$$\begin{aligned} \sum _{n=1}^{\infty }\frac{h_{j_r}(n)}{n^s}=\frac{2^s-2^{r+1}+1}{2^{s}-1}. \end{aligned}$$

Especially, for $s=r+1$, we have

$$\begin{aligned} \sum _{n=1}^{\infty }\frac{h_{j_r}(n)}{n^{r+1}}=\frac{1}{2^{r+1}-1}. \end{aligned}$$

(2.4)

Similarly, for reciprocal sum of Jordan’s function for $r>1$, and any real number $s>-r$, we can show that

$$\begin{aligned} \sum _{n=1}^{\infty }\frac{h_{1/J_r }(n)}{n^s}= \frac{2^{2r+s}-2^r(2^s+2)+1}{2^{2r+s}-2^{r+s}+1 }. \end{aligned}$$

For $s=1-r$, we have

$$\begin{aligned} \sum _{n=1}^{\infty }\frac{h_{1/J_r }(n)}{n^{1-r}}= \frac{1}{ 1-2^{r+1}}. \end{aligned}$$

(2.5)

3 Proofs

Proof of Theorem 1.1

By definition of $J_r(n)$ in (1.1) and considering Lemma 2.1, we have

$$\begin{aligned} \sum _{k\leqslant n}\log J_r(k)= & {} \sum _{k\leqslant n}\log \Big ( k^r\prod _{p|k}\Big (1-\frac{1}{p^r}\Big )\Big )\\= & {} r\sum _{k\leqslant n}\log k + \sum _{k\leqslant n}\sum _{p|k}\log \Big (1-\frac{1}{p^r}\Big )\\= & {} r\left( n\log n -n +\frac{1}{2}\log n + C + O\left( \frac{1}{n}\right) \right) + \sum _{p\leqslant n}\Big \lfloor \frac{n}{p}\Big \rfloor \log \Big (1-\frac{1}{p^r}\Big ). \end{aligned}$$

To approximate the last series, we write

$$\begin{aligned} \sum _{p\leqslant n}\Big \lfloor \frac{n}{p}\Big \rfloor \log \Big (1-\frac{1}{p^r}\Big )= & {} n\sum _{p\leqslant n}\frac{1}{p}\log \Big (1-\frac{1}{p^r}\Big )- \sum _{p\leqslant n}\Big \{\frac{n}{p}\Big \}\log \Big (1-\frac{1}{p^r}\Big )\nonumber \\= & {} n S_1(n) - S_2(n). \end{aligned}$$

(3.1)

To estimate $S_1(n)$, we can write

$$\begin{aligned} S_1(n)= \sum _{p\leqslant n}\frac{1}{p}\log \Big (1-\frac{1}{p^r}\Big )= & {} \sum _{p }\frac{1}{p}\log \Big (1-\frac{1}{p^r}\Big ) - \sum _{p> n}\frac{1}{p}\log \Big (1-\frac{1}{p^r}\Big )\\= & {} \alpha _r + O\Big (\frac{1}{n^r \log n}\Big ), \end{aligned}$$

where the last approximation is a special case of Lemma 2.3 and given that $\log (1-\frac{1}{p^r})\leqslant \frac{1}{p^r }$, so

$$\begin{aligned} \sum _{p> n}\frac{1}{p}\log \Big (1-\frac{1}{p^r}\Big ) \leqslant \sum _{p>n}\frac{1}{p^{r+1}}\ll \frac{1}{n^r \log n}. \end{aligned}$$

To approximate $S_2(n)$, we write

$$\begin{aligned} S_2(n)=\sum _{p\leqslant n}\Big \{\frac{n}{p}\Big \}\log \Big (1-\frac{1}{p^r}\Big )&\ll \sum _{p\leqslant n}\frac{1}{p^r} :=R(n)\ll {\left\{ \begin{array}{ll} \log \log n &{}if\quad r= 1\\ 1 &{} if \quad r>1. \end{array}\right. } \end{aligned}$$

Combining estimates $S_1(n)$ and $S_2(n)$ in (3.1), we get

$$\begin{aligned} \sum _{p\leqslant n}\Big \lfloor \frac{n}{p}\Big \rfloor \log \Big (1-\frac{1}{p^r}\Big )=n\alpha _r +R(n) \end{aligned}$$

and, therefore, we have

$$\begin{aligned} \sum _{k\leqslant n}\log J_r(k) = r n\log n +n(\alpha _r-r) +\frac{r}{2}\log n + R(n). \end{aligned}$$

This completes the proof of Theorem 1.1. $\square$

Proof of Corollary 1.2

By definition

$$\begin{aligned} G_{n}(J_r)= \left( \prod _{k=1}^{n}J_r(k)\right) ^{\frac{1}{n}}; \end{aligned}$$

by taking logarithms from both sides, we get

$$\begin{aligned} n \log G_{n}(J_r)= \sum _{k\leqslant n}\log J_r(k). \end{aligned}$$

Theorem 1.1 gives

$$\begin{aligned} \log G_{n}(J_r)=r \log n + (\alpha _r -r)+\frac{r}{2}\frac{\log n}{n} + O \Big ( \frac{R(n)}{n}\Big ). \end{aligned}$$

Now, using this fact $e^x=1+x+O(x^2)$ as $x\rightarrow 0$, we have

$$\begin{aligned} G_{n}(J_r)= & {} e^{r \log n}\, e^{(\alpha _r -r)}\, e^{\frac{r}{2}\frac{\log n}{n} + O \left( \frac{R(n)}{n}\right) }\\= & {} B_rn^r\left( 1+\frac{r}{2}\frac{\log n}{n} + O_r\Big ( \frac{R(n)}{n}\Big ) +O_r\Big (\frac{\log ^2 n}{n^2}\Big )\right) \\= & {} B_rn^r + \frac{r}{2}B_r n^{r-1}\log n +E(n), \end{aligned}$$

where the error term E(n) is $O(\log \log n)$ for $r = 1$ and $O(n^{r-1})$ for $r > 1$, which is the desired conclusion. $\square$

Proof of Theorem 1.3

First, by considering Euler’s infinite product factorization theorem, it is observed that

$$\begin{aligned} \beta _r = \sum _{d=1}^{\infty } \frac{\mu ^2(d)}{J_r(d) \,d^r}=\prod _p\left( 1+\frac{1}{p^r(p^r-1)}\right) , \end{aligned}$$

and

$$\begin{aligned} \beta '_r =\sum _{d=1}^{\infty } \frac{\mu ^2(d)}{J_r(d) \,d}= \prod _p\left( 1+\frac{1}{p(p^r-1)}\right) . \end{aligned}$$

Now, by considering Lemmas 2.5 and 2.2, we write

$$\begin{aligned} \sum _{n \leqslant x}\frac{1}{J_r(n)}= & {} \sum _{n\leqslant x}\frac{1}{n^r}\sum _{d|n}\frac{\mu ^2(d)}{J_r(d)}=\sum _{d\leqslant x} \frac{\mu ^2(d)}{J_r(d)}\sum _{\begin{array}{c} n\leqslant x \\ d|n \end{array}}\frac{1}{n^r} =\sum _{d\leqslant x} \frac{\mu ^2(d)}{J_r(d) d^r}\sum _{m\leqslant \frac{x}{d}}\frac{1}{m^r}\\= & {} \sum _{d\leqslant x} \frac{\mu ^2(d)}{J_r(d) d^r}\left( \frac{\left( \frac{x}{d}\right) ^{1-r}}{1-r}+\zeta (r)+O\left( \frac{1}{\left( \frac{x}{d}\right) ^r}\right) \right) \\= & {} \frac{1}{(1-r)x^{r-1}}\sum _{d\leqslant x} \frac{\mu ^2(d)}{J_r(d) d} +\zeta (r)\sum _{d\leqslant x} \frac{\mu ^2(d)}{J_r(d) d^r} +\sum _{d\leqslant x}O\left( \frac{\mu ^2(d)}{x^r J_r(d)}\right) , \end{aligned}$$

by Lemma 2.4 since $\frac{1}{\zeta (r)} n^r \leqslant J_r(n) \leqslant n^r$ for $r>1$, we see that $\sum _{d=1}^{\infty } \frac{\mu ^2(d)}{J_r(d) d}$ converges. So,

$$\begin{aligned} \sum _{n \leqslant x}\frac{1}{J_r(n)}= & {} \frac{1}{(1-r)x^{r-1}}\left( \sum _{d=1}^{\infty } \frac{\mu ^2(d)}{J_r(d) \,d}-\sum _{d>x} \frac{\mu ^2(d)}{J_r(d)\, d}\right) \\&+\zeta (r) \left( \sum _{d=1}^{\infty } \frac{\mu ^2(d)}{J_r(d) \,d^r}-\sum _{d>x} \frac{\mu ^2(d)}{J_r(d)\, d^r}\right) +\sum _{d\leqslant x} O\Big (\frac{\mu ^2(d)}{x^r J_r(d)}\Big )\\= & {} \zeta (r) \beta _r -\frac{\beta '_r}{(r-1)x^{r-1}}+ O\Big (\frac{1}{x^r}\Big ). \end{aligned}$$

This completes the proof of Theorem 1.3. $\square$

Proof of Theorem 1.4

By considering Lemma 2.6 and using the Abel’s summation formula, we get

$$\begin{aligned} \sum _{\begin{array}{c} k\leqslant y \\ (k,d)=1 \end{array}}\frac{\mu ^2(k)}{k^r}=\frac{r}{r-1}\frac{d}{\zeta (2)\psi (d)}-\frac{y^{r-1}d}{(r-1)\zeta (2)\psi (d)}+O\left( y^{1/2-r}2^{\omega (d)}\right) . \end{aligned}$$

Now, by considering Lemma 2.5, we have

$$\begin{aligned} \sum _{n\leqslant x}\frac{\mu ^2(n)}{J_r(n)}&= \sum _{n\leqslant x}\frac{\mu ^2(n)}{n^r}\sum _{d|n}\frac{\mu ^2(d)}{J_r(d)}\\&=\sum _{d\leqslant x}\frac{\mu ^2(d)}{d^rJ_r(d)}\sum _{\begin{array}{c} k\leqslant x/d \\ (k,d)=1 \end{array}}\frac{\mu ^2(k)}{k^r}\\&= \frac{r}{\zeta (2)(r-1)}\sum _{d\leqslant x}\frac{\mu ^2(d)}{d^{r-1}J_r(d)\psi (d)}-\frac{x^{1-r}}{\zeta (2)(r-1)}\sum _{d\leqslant x}\frac{\mu ^2(d)}{ J_r(d)\psi (d)}\\&\quad + O\left( x^{1/2-r} \sum _{d\leqslant x}\frac{\mu ^2(d) 2^{\omega (d)}}{d^{1/2}J_r(d)}\right) \\&= \frac{r\gamma _r}{r-1}-\frac{\gamma '_r x^{1-r}}{r-1}+O\left( x^{1/2-r}\right) , \end{aligned}$$

where

$$\begin{aligned} \gamma _r =\frac{1}{\zeta (2)}\sum _{d=1}^{\infty }\frac{\mu ^2(d)}{d^{r-1}J_r(d)\psi (d)}=\frac{1}{\zeta (2)}\prod _{p}\left( 1+\frac{1}{p^{r-1}(p^r-1)(p+1)}\right) , \end{aligned}$$

and

$$\begin{aligned} \gamma '_r =\frac{1}{\zeta (2)}\sum _{d=1}^{\infty }\frac{\mu ^2(d)}{J_r(d)\psi (d)}=\prod _p\left( 1-\frac{p^r-p}{p^2(p^r-1)}\right) . \end{aligned}$$

This completes the proof of Theorem 1.4. $\square$

Proof of Theorem 1.5

By considering (1.3) and (2.3), we write

$$\begin{aligned} \sum _{n\leqslant x}(-1)^{n-1}J_r(n)= & {} \sum _{d\leqslant x}h_{J_r}(d)\sum _{j\leqslant \frac{x}{d}}J_r(j)\\= & {} \sum _{d\leqslant x}h_{J_r}(d)\left( \frac{(\frac{x}{d})^{r+1}}{(r+1)\zeta (r+1)} +O\left( \frac{x}{d}\right) ^r\right) \\= & {} \frac{x^{r+1}}{(r+1)\zeta (r+1)}\sum _{d\leqslant x}\frac{h_{J_r}(d)}{d^{r+1}}+O\left( x^r\sum _{d\leqslant x}\frac{h_{J_r}(d)}{d^{r}}\right) \\= & {} \frac{x^{r+1}}{(r+1)\zeta (r+1)}\left( \sum _{d=1}^{\infty }\frac{h_{J_r}(d)}{d^{r+1}}-\sum _{d>x}\frac{h_{J_r}(d)}{d^{r+1}}+O(x^r)\right) \\= & {} \frac{x^{r+1}}{(2^{r+1}-1)(r+1)\zeta (r+1)}+O\left( x^r \right) . \end{aligned}$$

This completes the proof of Theorem 1.5. $\square$

Proof of Theorem 1.6

Proof of Theorem 1.6 using (2.3) and (2.5) and Theorem 1.3 is similar to the proof of Theorem 1.5. $\square$

Proof of Theorem 1.7

For $r>1$, by considering (1.5), write

$$\begin{aligned} \sum _{k=1}^{n}G_{k}(J_r)&= B_r\sum _{k=1}^{n}k^r+ \frac{r}{2}B_r\sum _{k=1}^{n}k^{r-1}\log k +O_r\left( \sum _{k=1}^{n}k^{r-1}\right) . \end{aligned}$$

We know

$$\begin{aligned} \sum _{k=1}^{n}k^r=\frac{n^{r+1}}{r+1}+O(n^{r}). \end{aligned}$$

Using Abel summation formula, we can write

$$\begin{aligned} \sum _{k=1}^{n}k^{r-1}\log k&= \left( n\log n -n +O(\log n)\right) n^{r-1}\\&\quad -(r-1)\int _{2}^{n}(t\log t -t +O(\log t))t^{r-2}\mathrm{d}t\\&=n^r\log n - n^r +O(n^{r-1}\log n)-(r-1)\left( \frac{n^r}{r} \log n-\frac{n^r}{r^2}\right) \\&\quad +(r-1)\left( \frac{n^r}{r}\right) +O\left( n^{r-1}\log n \right) \\&=\frac{1}{r}n^r\log n -\frac{1}{r^2}n^r+ O\left( n^{r-1}\log n\right) . \end{aligned}$$

So, we get

$$\begin{aligned} \sum _{k=1}^{n}G_{k}(J_r)=\frac{B_r}{r+1}n^{r+1}+\frac{B_r}{2}n^r\log n + O\left( n^r\right) . \end{aligned}$$

(3.2)

Also, we know

$$\begin{aligned} \sum _{k=1}^{n}J_r(k)= \frac{n^{r +1}}{(r+1)\zeta (r+1)} + O\Big ( n^{r} \Big ). \end{aligned}$$

(3.3)

Now, by considering (3.2) and (3.3), we have

$$\begin{aligned} C_{J_r}(n)&=\frac{\sum _{k=1}^{n}G_{k}(J_r)}{\sum _{k=1}^{n}J_r(k) }=\frac{\frac{B_r\,n^{r+1}}{r+1}+\frac{B_r}{2}n^r\log n + O\left( n^r\right) }{\frac{n^{r +1}}{(r+1)\zeta (r+1)} + O\left( n^{r} \right) }\\&= \frac{\frac{B_r}{r+1}n^{r+1}\left( 1+\frac{r+1}{2}\frac{\log n}{n}+O\left( \frac{1}{n}\right) \right) }{\frac{n^{r +1}}{(r+1)\zeta (r+1)} \left( 1+O\left( \frac{1}{n}\right) \right) }\\&=B_r\zeta (1+r)\left( 1+\frac{r+1}{2}\frac{\log n}{n}+O\left( \frac{1}{n}\right) \right) \left( 1+O\left( \frac{1}{n}\right) \right) \\&=B_r\zeta (1+r)+\frac{B_r(r+1)}{2}\frac{\log n}{n}+O\left( \frac{1}{n}\right) . \end{aligned}$$

This completes the proof of Theorem 1.7. $\square$

Proof of Lemma 2.5

Suppose $n= p_1^{\alpha _1} p_2^{\alpha _2}\dots p_k^{\alpha _k}$, where $\alpha _i>0$. Then since the right-hand side of (2.1) is multiplicative,

$$\begin{aligned} \sum _{d|n}\frac{\mu ^2(d)}{J_r(d)}= & {} \prod _{i=1}^{k}\sum _{j=0}^{a_i}\frac{\mu ^2(p_i^{j})}{J_r(p_i^{j})}\\= & {} \prod _{i=1}^{k}\left( 1+\frac{1}{p_i^r-1}\right) =\prod _{i=1}^{k} \frac{1}{1-\frac{1}{p_i^r}}=\frac{n^r}{J_r(n)}. \end{aligned}$$

This completes the proof of Lemma 2.5. $\square$