# On a Sum Involving the Sum-of-Divisors Function

Authors: Feng Zhao; Jie Wu
Journal: Journal of Mathematics
 (2021)

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## Abstract

Letσn be the sum of all divisors of n and let t be the integral part of t. In this paper, we shall prove that ∑n≤xσx/n=π2/6xlogx+Oxlogx2/3log2x4/3 for x⟶∞, and that the error term of this asymptotic formula is Ωx.


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## Body

## 1. Introduction

As usual, denote byφn the Euler function and by t the integral part of real t, respectively. Recently, Bordellès et al. [1] studied the asymptotic behaviour of the quantity(1)Sφx≔∑n≤xφxn,for x⟶∞. By exponential sum technique, they proved that(2)26294009⋅6π2+o1xlogx≤Sφx≤26294009⋅6π2+13804009+o1xlogx,and conjectured that(3)Sφx∼6π2xlogx,asx⟶∞.Very recently, Wu [2] improved (2) and Zhai [3] resolved conjecture (3) by showing(4)Sφx=6π2xlogx+Oxlogx2/3log2x1/3,and also proved that the error term in (4) is Ωx, where log2 denotes the iterated logarithm. Some related works can be found in [4, 5]. Since the sum-of-divisors function σn≔∑d|nd has similar properties as the Euler function φn in many cases, it seems natural and interesting to consider its analogy of (3).Our result is as follows.Theorem 1.
(i)
Forx⟶∞, we have(5)Sσx≔∑n≤xσxn=π26xlogx+Oxlogx2/3log2x4/3.(ii)
LetEx be the error term in (5). Then, for x⟶∞, we have(6)Ex=Ωx,i.e.limsupx⟶∞Exx>0.Letμn be the Möbius function and define idn=n and 1n=1 for all integers n≥1. Then, φ=id∗μ and σ=id∗1. In Zhai’s approach proving (4), the inequality(7)∑n≤xμn≪xexp−clogx,x≥1,plays a key role, where c>0 is a positive constant. Clearly, such a bound is not true for 1. By refining Zhai’s approach, we shall prove our result.


## 2. Preliminary Lemmas

As in [3], we need some bounds on exponential sums of the type ∑N≤n<N′eT/n where N<N′≤2N. For large values of N, Zhai used the theory of exponent pair, and for smaller ones the Vinogradov method. Both estimates are contained in the following general theorem of Karatsuba [6, Theorem 1], which will be a key tool for proving Theorem 1.Lemma 1.
Letk≥2 and M and P be integers, P being positive. Let f∈Ck+1M,M+P;ℝ. Suppose that there exist positive absolute constants c0,c1,c2,c3, and c4 such that c0<1,c1<1, and c2+c4<c1; an integer r such that c0k≤r≤k; and distinct numbers sj≥2j=1,…,r not exceeding k, such that for M≤t≤M+P the following inequalities are satisfied:(i)
fk+1t/k+1!≤P−c1k+1.(ii)
P−c2sj≤fsjt/sj!≤P−c3sj,j=1,…,r.
Then, for each positive integerP1 not exceeding P, we have(8)∑M≤m≤M+P1−1efm≤AP1−c/k2,where et≔e2πit and A>0, c>0 are absolute constants.The next two lemmas are essentially a special case of [7, Lemmas 2.5 and 2.6] with a=1. The only difference is that the ranges of T and N here are slightly larger than those of [7, Lemmas 2.5 and 2.6] (T≥N2 in place of T≥N3/2 and N≤x2/3 in place of N≤x1/2, respectively). Although the proof is completely similar, for the convenience of readers, we still reproduce a proof here.Lemma 2.
Lete100≤N<N′≤2N and T≥N3/2. Then, there exists an absolute positive constant c5 such that(9)∑N≤n<N′eTn≪Nexp−c5log3Nlog2T,where the implied constant is absolute.Proof.
We apply Lemma1 to ft≔T/t with M=N,P=N,P1=N′−N. For this, we choose(10)c0=1100,c1=99100,c2=87100,c3=34,c4=1100,k=100logT/nlogN,and take the sj to be all integers s such that(11)4logT/nlogN≤s≤5logT/nlogN⋅
Obviously the numberr of sj is between c0k and k. Next we shall verify that ft satisfies the conditions (i) and (ii) of Lemma 1 with the parameters chosen above.
ForN≤t≤2N, we have(12)fk+1tk+1!=Tt−k−2≤TN−k−2=N−η1,where(13)η1≔k+1−logT/nlogN≥k+1−1100k≥99100k+1=c1k+1.
Similarly forN≤t≤2N, we find the inequality fsjt/sj!≤N−η3, where(14)η3≔sj−logT/nlogN≥34sj=c3sj.
For the lower bound of (ii), we have(15)fsjtsj!=Tt−1−sj≥T2N−1−sj=N−η2,where(16)η2≔sj−logT/nlogN+log2logNsj+1≤45sj+log2100sj+1≤87100sj=c2sj.
From Lemma1, there exist two positive constants c and A such that(17)∑N≤n<N′eTn≤AN1−c/k2≤ANexp−c5log3Nlog2T/n,with c5≔10−4c. This completes the proof of Lemma 2.Lemma 3.
Defineψt≔t−t−1/2. Let c5 be the constant defined by Lemma 2 and c6≔8/92c5,c∗≔3/5c6−1/3. Then, we have(18)∑N≤n<N′1nψxn≪e−c6logN3/logx2logN3logx2,uniformly for x≥10, expc∗logx2/3≤N≤x2/3 and N<N′≤2N.Proof.
By invoking a classical result onψt (see 8, page 39]) we can write, for any H≥1,(19)∑N≤n<N′ψxn≪NH−1+∑1≤h≤Hh−1∑N≤n<N′ehxn.
An application of Lemma2 with T=hx≥x≥N3/2 yields(20)∑N≤n<N′ψxn≪NH−1+e−c5logN3/log2HxlogH.
TakingH=explogN3/logx2≤x8/27, we easily deduce that(21)∑N≤n<N′ψxn≪Ne−logN3/logx2+e−c6logN3/logx2logN3logx2.
The first term can be absorbed by the second, sincec5 can be chosen small enough to ensure that c6<1 and since expc∗logx2/3≤N implies logN3/logx2≥c∗. Hence,(22)∑N≤n<N′ψxn≪Ne−c6logN3/logx2logN3logx2,and an Abel summation produces the required result.Lemma 4.
Let2≤z1<z2≤x and Fxt≔1/tψx/t. Denote by VFxz1,z2 the total variation of Fx on z1,z2. Then,(23)VFxz1,z2≪xz12+1z1,where the implied constant is absolute.Proof.
Ifz1=t0<t1<⋯<tn=z2 is a partition of the interval z1,z2, then(24)∑k=1nFxtk−Fxtk−1=∑k=1n1tkψxtk−1tk−1ψxtk−1≤∑k=1n1tk−1tk−1ψxtk+∑k=1n1tk−1ψxtk−ψxtk−1.
Sinceψt≤1 for all t, we have(25)∑k=1n1tk−1tk−1ψxtk≤1z1−1z2<1z1⋅
On the other hand, sinceψu is of period 1, we have(26)∑k=1n1tk−1ψxtk−ψxtk−1≤1z1xz1+1Vψ0,1≤2xz12+1z1.
Inserting these two bounds into (24), we obtain the required result.


## 3. Proof of Theorem1

### 3.1. A Formula on the Mean Value ofσn

Lemma 5.
(i)
Forx≥2 and 1≤z≤x1/3, we have(27)∑n≤xσn=π212x2−xz−z2+z2z+Oxz−Δx,z,where(28)Δx,z≔∑d≤x/zxdψxd.(ii)
Forx⟶∞, we have(29)∑n≤xσn=π212x2+Oxlogx.Proof.
Usingσn=∑dm=nm, the hyperbole principle of Dirichlet allows us to write(30)∑n≤xσn=∑dm≤xm=S1+S2−S3,where(31)S1≔∑d≤x/z∑m≤x/dm,S2≔∑m≤z∑d≤x/mm,S3≔∑d≤x/z∑m≤zm.
Firstly we have(32)S2=∑m≤zmxm=xz+Oz2,(33)S3=xzzz+12=xzz+12z+Oz2.
Secondly we can write(34)S1=12∑d≤x/zxd−ψxd−12xd−ψxd+12=12∑d≤x/zx2d2−2xdψxd+ψxd2−14=π212x2−12xz−Δx,z+Ox/z,where Δx,z is as in (28). Inserting (32), (33), and (34) into (30) and using z2≤x/z, we get (27).
Takingz=1 in (27) and noticing that(35)∑d≤x1d2=π26+O1x,∑d≤xxdψxd≪xlogx,we obtain the required bound. This completes the proof.


### 3.2. Estimates of Error Terms

Lemma 6.
LetN0≔exp6/c6logx2/3log2x1/3, where c6 is given as in Lemma 3. Let Δx,z be defined by (28). Then, for x≥10 and 2≤z≤N0, we have(36)∑N0<n≤xΔxn,z+∑N0<n≤xΔxn−1,z≪x1logx3+logxz.Proof.
Denote byΔ1x,z and Δ2x,z two sums on the left-hand side of (36), respectively. By (28) of Lemma 5, we can write(37)Δ1x,z=x∑N0<n≤x∑d≤x/nz1dnψxdn=x∑d≤x/N0z1d∑N0<n≤minx,x/dz1nψxdn=xΔ1†x,z+xΔ1♯x,z,where(38)Δ1†x,z≔∑d≤x/N0z1d∑N0<n≤x/d2/31nψxdn,Δ1♯x,z≔∑d≤x/N0z1d∑x/d2/3<n≤minx,x/dz1nψxdn.
For0≤k≤logx/d2/3/N0/log2, let Nk≔2kN0 and define(39)Skd≔∑Nk<n≤2Nk1nψxdn.
Noticing thatN0≤Nk≤x/d2/3, we can apply Lemma 3 to derive that(40)Skd≪e−ϑlogNk3/logx/d2,with ϑt≔c6t−logt. It is clear that ϑt is increasing on c6,∞. On the other hand, for k≥0 and d≥1, we have(41)logNk3/logx/d2≥logN03/logx2=6/c6log2x.
Thus,(42)ϑlogNk3logx/d2≥ϑ6c6log2x=6log2x−log6c6log2x≥5log2x,which implies that Skd≪logx−5. Inserting this into the expression of Δ1†x,z, we get(43)Δ1†x,z≪∑d≤x/N0z1d∑2kN0≤x/d2/3Skd≪logx−3.
Next we boundΔ1♯x,z. Let Ft be a function of bounded variation on n,n+1 for each integer n and let VFn,n+1 be the total variation of F on n,n+1. Integrating by parts, we have(44)∫nn+1t−n−12dFt=12Fn+1+Fn−∫nn+1Ftdt.
From this, we can derive that(45)12Fn+1+Fn=∫nn+1Ftdt+OVFn,n+1,for n≥1. Summing over n, we find that(46)∑N1<n≤N2Fn=∫N1N2Ftdt+12FN1+FN2+OVFN1,N2.
We apply this formula to(47)Fx/dt=1tψx/dt,N1=x/d2/3,N2=minx,xdz.
According to Lemma4, we have VFx/dN1,N2≪x/d−1/3, and thus by putting u=x/d/t, we obtain, with the notation xd,1=maxx/d,z and xd,2=x/d1/3,(48)∑x/d2/3<n≤minx,x/dz1nψxdn=∫xd,1xd,2ψuudu+Oxd−1/3≪z−1+xd−1/3≪z−1,where we have used the fact that z≤N0 and d≤x/N0z⇒z≤x/d1/3 and the bound(49)∫xd,1xd,2ψuudu=∫xd,1xd,2∫xd,1u​ψtdtduu2−1xd,22/3∫xd,1xd,2ψtdt≪xd,1−1+xd,2−2/3≪z−1+x/d−2/3≪z−1.
Using (48), a simple partial integration allows us to derive that(50)Δ1♯x,z≪z−1∑d≤x/N0zd−1≪z−1logx.
Combining (43) and (50), it follows that(51)Δ1x,z≪xlogx−3+xz−1logx.
Similarly, we can prove the same bound forΔ2x,z. This completes the proof.


### 3.3. End of the Proof of Theorem 1

Letc6 be the constant given as in Lemma 3 and N0≔exp6/c6logx2/3log2x1/3. Let z∈2,N0 be a parameter to be chosen later.Puttingd=x/n, we have x/n−1<d≤x/n and x/d+1<n≤x/d. We have, with the convention σ0=0,(52)Sσx=∑d≤xσd∑x/d+1<n≤x/d1=∑dn≤xσd−∑dn≤x,d≥2σd−1=∑dn≤xσd−σd−1.By the hyperbole principle of Dirichlet, we can write(53)Sσx=S1x,σ+S2x,σ−S3x,σ,where(54)S1x,σ≔∑d≤x,dn≤xσd−σd−1,S2x,σ≔∑n≤x,dn≤xσd−σd−1,S3x,σ≔∑d≤x,n≤xσd−σd−1.With the help of the boundσn≪nlog2n, we can derive that(55)S3x,σ=xσx≪xlog2x.For evaluatingS1x,σ, we write(56)S1x,σ=∑d≤xσd−σd−1xd=x∑d≤xσd−σd−1d+O∑d≤xσd−σd−1.With the help of Lemma5 (ii), a simple partial integration gives us(57)∑d≤xσd−σd−1d=∑d≤xσddd+1=∑d≤xσdd2−∑d≤xσdd2d+1=∫1−xt−2dπ212t2+Otlogt+O1=π212logx+O1,∑d≤xσd−σd−1≪∑d≤xσd≪x.Inserting these estimates into (56), we find that(58)S1x,σ=π212xlogx+Ox.Finally, we evaluateS2x,σ. For this, we write(59)S2x,σ=S2†x,σ+S2♯x,σ,where(60)S2†x,σ≔∑n≤N0,dn≤xσd−σd−1,S2♯x,σ≔∑N0<n≤x,dn≤xσd−σd−1.By the bound thatσn≪nlog2n, we have(61)S2†x,σ=∑n≤N0σxn≪∑n≤N0xnlog2x≪xlogx2/3log2x4/3.On the other hand, (27) of Lemma 5 allows us to derive that(62)∑d≤xσd−∑d≤x−1σd=π212x2−x−12−Δx,z+Δx−1,z+Oxz=π26x−Δx,z+Δx−1,z+Oxz,where Δx,z is given by (28). Thus,(63)S2♯x,σ=∑N0<n≤xπ26⋅xn−Δxn,z+Δxn−1,z+Oxnz=π212xlogx+Oxlogx2/3log2x1/3+xz−1logx−Δ1x,z+Δ2x,z,where(64)Δ1x,z≔∑N0<n≤xΔxn,z≪xlogx−3+xz−1logx,Δ2x,z≔∑N0<n≤xΔxn−1,z≪xlogx−3+xz−1logx,thanks to Lemma 6. Inserting these estimates into (59), we find that(65)S2x,σ=π212xlogx+Oxlogx2/3log2x4/3+xz−1logx,Now (5) follows from (53), (55), (58), and (66) with the choice of z=logx1/3.


### 3.4. Proof of Theorem 1

(ii) For any odd primep, (52) allows us to write(66)∑d∣pσd−σd−1=Sσp−Sσp−1=π26logp−logp−1+Ep−Ep−1≥Ep−Ep−1≥−2E∗p,where E∗p≔maxEp,Ep−1. On the other hand, we have(67)∑d|pσd−σd−1=σp−σp−1+1≤p+1−p−1+12p−1+2+1+1≤−14p.Thus,E∗p≥1/8p for all odd primes.


## 3.1. A Formula on the Mean Value ofσn

Lemma 5.
(i)
Forx≥2 and 1≤z≤x1/3, we have(27)∑n≤xσn=π212x2−xz−z2+z2z+Oxz−Δx,z,where(28)Δx,z≔∑d≤x/zxdψxd.(ii)
Forx⟶∞, we have(29)∑n≤xσn=π212x2+Oxlogx.Proof.
Usingσn=∑dm=nm, the hyperbole principle of Dirichlet allows us to write(30)∑n≤xσn=∑dm≤xm=S1+S2−S3,where(31)S1≔∑d≤x/z∑m≤x/dm,S2≔∑m≤z∑d≤x/mm,S3≔∑d≤x/z∑m≤zm.
Firstly we have(32)S2=∑m≤zmxm=xz+Oz2,(33)S3=xzzz+12=xzz+12z+Oz2.
Secondly we can write(34)S1=12∑d≤x/zxd−ψxd−12xd−ψxd+12=12∑d≤x/zx2d2−2xdψxd+ψxd2−14=π212x2−12xz−Δx,z+Ox/z,where Δx,z is as in (28). Inserting (32), (33), and (34) into (30) and using z2≤x/z, we get (27).
Takingz=1 in (27) and noticing that(35)∑d≤x1d2=π26+O1x,∑d≤xxdψxd≪xlogx,we obtain the required bound. This completes the proof.


## 3.2. Estimates of Error Terms

Lemma 6.
LetN0≔exp6/c6logx2/3log2x1/3, where c6 is given as in Lemma 3. Let Δx,z be defined by (28). Then, for x≥10 and 2≤z≤N0, we have(36)∑N0<n≤xΔxn,z+∑N0<n≤xΔxn−1,z≪x1logx3+logxz.Proof.
Denote byΔ1x,z and Δ2x,z two sums on the left-hand side of (36), respectively. By (28) of Lemma 5, we can write(37)Δ1x,z=x∑N0<n≤x∑d≤x/nz1dnψxdn=x∑d≤x/N0z1d∑N0<n≤minx,x/dz1nψxdn=xΔ1†x,z+xΔ1♯x,z,where(38)Δ1†x,z≔∑d≤x/N0z1d∑N0<n≤x/d2/31nψxdn,Δ1♯x,z≔∑d≤x/N0z1d∑x/d2/3<n≤minx,x/dz1nψxdn.
For0≤k≤logx/d2/3/N0/log2, let Nk≔2kN0 and define(39)Skd≔∑Nk<n≤2Nk1nψxdn.
Noticing thatN0≤Nk≤x/d2/3, we can apply Lemma 3 to derive that(40)Skd≪e−ϑlogNk3/logx/d2,with ϑt≔c6t−logt. It is clear that ϑt is increasing on c6,∞. On the other hand, for k≥0 and d≥1, we have(41)logNk3/logx/d2≥logN03/logx2=6/c6log2x.
Thus,(42)ϑlogNk3logx/d2≥ϑ6c6log2x=6log2x−log6c6log2x≥5log2x,which implies that Skd≪logx−5. Inserting this into the expression of Δ1†x,z, we get(43)Δ1†x,z≪∑d≤x/N0z1d∑2kN0≤x/d2/3Skd≪logx−3.
Next we boundΔ1♯x,z. Let Ft be a function of bounded variation on n,n+1 for each integer n and let VFn,n+1 be the total variation of F on n,n+1. Integrating by parts, we have(44)∫nn+1t−n−12dFt=12Fn+1+Fn−∫nn+1Ftdt.
From this, we can derive that(45)12Fn+1+Fn=∫nn+1Ftdt+OVFn,n+1,for n≥1. Summing over n, we find that(46)∑N1<n≤N2Fn=∫N1N2Ftdt+12FN1+FN2+OVFN1,N2.
We apply this formula to(47)Fx/dt=1tψx/dt,N1=x/d2/3,N2=minx,xdz.
According to Lemma4, we have VFx/dN1,N2≪x/d−1/3, and thus by putting u=x/d/t, we obtain, with the notation xd,1=maxx/d,z and xd,2=x/d1/3,(48)∑x/d2/3<n≤minx,x/dz1nψxdn=∫xd,1xd,2ψuudu+Oxd−1/3≪z−1+xd−1/3≪z−1,where we have used the fact that z≤N0 and d≤x/N0z⇒z≤x/d1/3 and the bound(49)∫xd,1xd,2ψuudu=∫xd,1xd,2∫xd,1u​ψtdtduu2−1xd,22/3∫xd,1xd,2ψtdt≪xd,1−1+xd,2−2/3≪z−1+x/d−2/3≪z−1.
Using (48), a simple partial integration allows us to derive that(50)Δ1♯x,z≪z−1∑d≤x/N0zd−1≪z−1logx.
Combining (43) and (50), it follows that(51)Δ1x,z≪xlogx−3+xz−1logx.
Similarly, we can prove the same bound forΔ2x,z. This completes the proof.


## 3.3. End of the Proof of Theorem 1

Letc6 be the constant given as in Lemma 3 and N0≔exp6/c6logx2/3log2x1/3. Let z∈2,N0 be a parameter to be chosen later.Puttingd=x/n, we have x/n−1<d≤x/n and x/d+1<n≤x/d. We have, with the convention σ0=0,(52)Sσx=∑d≤xσd∑x/d+1<n≤x/d1=∑dn≤xσd−∑dn≤x,d≥2σd−1=∑dn≤xσd−σd−1.By the hyperbole principle of Dirichlet, we can write(53)Sσx=S1x,σ+S2x,σ−S3x,σ,where(54)S1x,σ≔∑d≤x,dn≤xσd−σd−1,S2x,σ≔∑n≤x,dn≤xσd−σd−1,S3x,σ≔∑d≤x,n≤xσd−σd−1.With the help of the boundσn≪nlog2n, we can derive that(55)S3x,σ=xσx≪xlog2x.For evaluatingS1x,σ, we write(56)S1x,σ=∑d≤xσd−σd−1xd=x∑d≤xσd−σd−1d+O∑d≤xσd−σd−1.With the help of Lemma5 (ii), a simple partial integration gives us(57)∑d≤xσd−σd−1d=∑d≤xσddd+1=∑d≤xσdd2−∑d≤xσdd2d+1=∫1−xt−2dπ212t2+Otlogt+O1=π212logx+O1,∑d≤xσd−σd−1≪∑d≤xσd≪x.Inserting these estimates into (56), we find that(58)S1x,σ=π212xlogx+Ox.Finally, we evaluateS2x,σ. For this, we write(59)S2x,σ=S2†x,σ+S2♯x,σ,where(60)S2†x,σ≔∑n≤N0,dn≤xσd−σd−1,S2♯x,σ≔∑N0<n≤x,dn≤xσd−σd−1.By the bound thatσn≪nlog2n, we have(61)S2†x,σ=∑n≤N0σxn≪∑n≤N0xnlog2x≪xlogx2/3log2x4/3.On the other hand, (27) of Lemma 5 allows us to derive that(62)∑d≤xσd−∑d≤x−1σd=π212x2−x−12−Δx,z+Δx−1,z+Oxz=π26x−Δx,z+Δx−1,z+Oxz,where Δx,z is given by (28). Thus,(63)S2♯x,σ=∑N0<n≤xπ26⋅xn−Δxn,z+Δxn−1,z+Oxnz=π212xlogx+Oxlogx2/3log2x1/3+xz−1logx−Δ1x,z+Δ2x,z,where(64)Δ1x,z≔∑N0<n≤xΔxn,z≪xlogx−3+xz−1logx,Δ2x,z≔∑N0<n≤xΔxn−1,z≪xlogx−3+xz−1logx,thanks to Lemma 6. Inserting these estimates into (59), we find that(65)S2x,σ=π212xlogx+Oxlogx2/3log2x4/3+xz−1logx,Now (5) follows from (53), (55), (58), and (66) with the choice of z=logx1/3.


## 3.4. Proof of Theorem 1

(ii) For any odd primep, (52) allows us to write(66)∑d∣pσd−σd−1=Sσp−Sσp−1=π26logp−logp−1+Ep−Ep−1≥Ep−Ep−1≥−2E∗p,where E∗p≔maxEp,Ep−1. On the other hand, we have(67)∑d|pσd−σd−1=σp−σp−1+1≤p+1−p−1+12p−1+2+1+1≤−14p.Thus,E∗p≥1/8p for all odd primes.



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