Lambert series in analytic number theory

Jordan Bell
[email protected]
(February 22, 2023)

Tour of 19th and early 20th century analytic number theory.

1 Introduction

Let d(n) denote the number of positive divisors of n. For |z|<1,


2 Euler

The first use of the term “Lambert series” was by Euler to describe the roots of an equation.

Euler writes in E25 [28] about the particular value of a Lambert series.

3 Lambert

Bullynck [7, pp. 157–158]: “As he recorded in his scientific diary, the Monatsbuch, Lambert started thinking about the divisors of integers in June 1756. An essay by G.W. Krafft (1701–1754) in the St. Petersburg Novi Commentarii seems to have triggered Lambert’s interest [Bopp 1916, p. 17, 40].”

Bullynck [7, p. 163]:

Lambert did more than deliver the factor table. He also addressed the absence of any coherent theory of prime numbers and divisors. Filling such a lacuna could be important for the discovery of new and more primality criteria and factoring tests. For Lambert the absence of such a theory was also an occasion to apply the principles laid out in his philosophical work. A fragmentary theory, or one with gaps, needed philosophical and mathematical efforts to mature.

To this aim [prime recognition] and others I have looked into the theory of prime numbers, but only found certain isolated pieces, which did not seem possible to make easily into a connected and well formed system. Euclid has few, Fermat some mostly unproven theorems, Euler individual fragments, that anyway are farther away from the first beginnings, and leave gaps between them and the beginnings. [Lambert 1770, p. 20]

Bullynck [7, pp. 164–165]:

In 1770, Lambert presented two sketches of what would be needed for something like a theory of numbers. The first dealt mainly with factoring methods [Lambert 1765-1772, II, pp. 1–41], while the second gave a more axiomatic treatment [Lambert 1770, pp. 20–48]. In the first essay, Lambert explained how, for composite number with small factors, Eratosthenes’ sieve could be used and optimised. For larger factors, Lambert explained that approximation from above, starting by division by numbers that are close to the square root of the tested number p, was more advantageous. For both methods, Lambert advised the use of tables. The second essay had more theoretical bearings. Lambert rephrased Euclid’s theorems for use in factoring, included the greatest common divisor algorithm, and put the idea of relatively prime numbers to good use. He also noted that binary notation, because of the frequent symmetries, could be helpful. Finally,Lambert also recognized Fermat’s little theorem as a good, though not infallible criterion for primality, “but the negative example is very rar” [Lambert 1770, p. 43].

Monatsbuch, September 1764, “Singula haec in Capp. ult. Ontol. occurunt”, and Anm. 5, Anm. 25, 1764, Anm. 12 1765, Anm. 19, 1765 [2].

Lambert [53, pp. 506–511, §875]

Youschkevitch [87]

Lorey [59, p. 23]

Löwenhaupt [60, p. 32]

4 Krafft

Krafft [50, pp. 244–245]

5 Servois

Servois [72] and [73, p. 166]

6 Lacroix

Lacroix [51, pp. 465–466, §1195]

7 Klügel

Klügel [46, pp. 52–53, s.v. “Theiler einer Zahl”, §12]:

Ist N=αmβnγp, wo α,β,γ, Primzahlen sind; so erhellet auch leicht, daßalle Theiler von N, die Einheit und die Zahl selbst mit engeschlossen, durch die Glieder des Products


argestelle werden. Die Anzahl der Glieder dieses Products, d. i. die Anzahl aller Theiler von N, ist offenbar =(m+1)(n+1)(p+1). Für das obige Beispiel =432=24, wo die Einheit mit engeschlossen ist.
In der aus der Entwickelung von


entspringenden Reihe:


welche Lambert in seiner Architektonik S. 507. mittheilt, enthalt jeder Coefficient so viele Einheiten, als der Exponent der entsprechendenden Potenz von x Theiler hat.

8 Stern

Stern [77]

9 Clausen

Clausen [21] states that


and that the right-hand series converges quickly for small x. Clausen does prove this expansion, and a proof is later given by Scherk [68]. Scherk’s argument uses the fact


We write


The series is


We sum the terms in the first row and column: the sum of these is


Then, from what remains we sum the terms in the second row and column: the sum of these is


Then, from what remains, we sum the terms in the third row and column: the sum of these is



10 Eisenstein

Eisenstein [27] states that for |z|<1,


For t=1z, Eisenstein states that


is equal to


Expressing Lambert series using continued fractions is relevant to the irrationality of the value of the series. See Borwein [3]. See also Zudilin [90].

11 Möbius

Möbius [62]

12 Jacobi

Jacobi’s Fundamenta nova [44, §40, 66 and p. 185]

Chandrasekharan [20, Chapter X]: using Lambert series to prove the four squares theorem.

13 Dirichlet

Dirichlet [25]

Fischer [29]

14 Cauchy

Cauchy [13] and [14] two memoirs in the same volume.

15 Burhenne

Burhenne [8] says the following about Lambert series. For


we have




so that


It is apparent that if k>n, then




The above suggests finding explicit expressions for Fk(n)(0). Burhenne cites Sohncke [74, pp. 32–33]: for even k,

dn(xpxk-ak)dxn =(-1)nn!kak-p-1(1(x-a)n+1-(-1)p1(x+a)n+1)

and for odd k,

dn(xpxk-ak)dxn =(-1)nn!kak-p-11(x-a)n+1



For a=1 and x=0,


from which


and thus

cos(2h(k+1)πk+(n+1)ϕh) =cos(2h(k+1)πk+(n+1)(π-2hπk))

For even k, taking p=k we have




For odd k, taking p=k we have




Using the identity, for h2π,


we get for even k,

Fk(n)(0) ={n!kcotnπksinnπknn!k(1-(-1)n+1)+2n!k(12k-1)k|n

For odd k,

Fk(n)(0) ={n!kcscnπksinnπknn!k+2n!kk-12k|n.

16 Zehfuss

Zehfuss [88]

17 Bernoulli numbers

The Bernoulli polynomials are defined by


The Bernoulli numbers are defined by Bm=Bm(0).

We denote by [x] the greatest integer x, and we define {x}=x-[x], namely, the fractional part of x. We define Pm(x)=Bm({x}), the periodic Bernoulli functions.

18 Euler-Maclaurin summation formula

Euler E47 and E212, §142, for the summation formula. Euler’s studies the gamma function in E368. In particular, in §12 he gives Stirling’s formula, and in §14 he obtains Γ(1)=-γ. Euler in §142 of E212 states that


Bromwich [6, Chapter XII]

The Euler-Maclaurin summation formula [5, p. 280, Ch. VI, Eq. 35] tells us that for fC([0,1]),




Poisson and Jacobi on the Euler-Maclaurin summation formula.

19 Schlömilch

Schlömilch [69] and [71, p. 238], [70]

For m1,

0t2m-1e2πt-1𝑑t=(-1)m+1B2m4m. (1)

For α>0,

0sinαte2πt-1𝑑t=14+12(1eα-1-1α) (2)


01-cosαte2πt-1dtt=14α+12(log(1-e-α)-logα). (3)

For ξ>0 and n1, using (2) with α=ξ,2ξ,3ξ,,2nξ and also using


we get

m=12n(1emξ-1-1mξ) =m=12n(-12+20sinmξte2πt-1𝑑t)

Using cos(a+b)=cosacosb-sinasinb, this becomes

m=12n(1emξ-1-1mξ) =-n+01e2πt-1(1-cos2nξt)cotξt2dt (4)

For α=2nξ, (3) tells us



log2nξ=n+log(1-e-2nξ)-logξξ-2ξ01-cos2nξte2πt-1dtt (5)

Adding (4) and (5) gives




and using (2) this becomes


We write


and we shall obtain an asymptotic formula for I2n(ξ).

We apply the Euler-Maclaurin summation formula. Let h>0, and for f(t)=cosht we have f(t)=-hsinht, and for m1 we have f(2m)(t)=(-1)mh2mcosht and f(2m-1)(t)=(-1)mh2m-1sinht. Thus the Euler-Maclaurin formula yields


Using the identity cotθ2=1+cosθsinθ and dividing by sinh, this becomes

12coth2=1h+m=1k1(2m)!B2m(-1)mh2m-1+1sinhR2k. (6)

Because Pm(1-η)=Pm(η) for even m,

R2k =-01P2k(η)(2k)!(-1)kh2kcoshηdη

Since P2k(η)-B2k does not change sign on (0,1), by the mean-value theorem for integration there is some θ=θ(h,k), 0<θ<1, such that (using 01P2k(η)𝑑η=0)


Therefore (6) becomes

12coth2-1h =m=1k1(2m)!B2m(-1)mh2m-1





We apply the above to I2n(ξ), and get, for any k1,

I2n(ξ) =20(Ek(ξt)-m=1k-11(2m)!B2m(-1)m(ξt)2m-1)1-cos2nξte2πt-1𝑑t

Using (1),

0t2m-11-cos2nξte2πt-1𝑑t =0t2m-1e2πt-1𝑑t-0t2m-1cos2nξte2πt-1𝑑t



By (2),


For m1,


which for x=2nξ becomes




Thus I2n(ξ) is

I2n(ξ) =-m=1k-11(2m)!B2m(-1)mξ2m-1((-1)m+1B2m2m+(-1)mf(2m-1)(2nξ))


|0Ek(ξt)1-cos2nξte2πt-1𝑑t| =|0(-1)k+1(ξt)2k(2k)!sinξtB2kcosξtθ1-cos2nξte2πt-1dt|

It is a fact that for all u,


we obtain



I2n(ξ) =m=1k-1B2m2(2m)!2mξ2m-1-m=1k-1B2m(2m)!ξ2m-1f(2m-1)(2nξ)

Therefore we have


Taking n,


20 Voronoi summation formula

The Voronoi summation formula [22, p. 182] states that if f: is a Schwartz function, then

n=1d(n)f(n) =0f(t)(logt+2γ)𝑑t+f(0)4

where K0 and Y0 are Bessel functions.

Let 0<x<1. For f(t)=e-tx, we compute




the exponential integral. Then the Voronoi summation formula yields


Egger and Steiner [26] give a proof of the Voronoi summation formula involving Lambert series.

Kluyver [47] and [48]

Guinand [36]

21 Curtze

Curtze [23]

22 Laguerre

Laguerre [52]

23 V. A. Lebesgue

V. A. Lebesgue [56]:

24 Bouniakowsky

Bouniakowsky [4]

25 Chebyshev

Chebyshev [80]

26 Catalan

Catalan [9]

Catalan [10, p. 89]

Catalan [11, p. 119, §CXXIV] and [12, pp. 38–39, §CCXXVI]

27 Pincherle

Pincherle [63]

28 Glaisher

Glaisher [34, p. 163]

29 Günther

Günther [37, p. 83] and [38, p. 178]

30 Stieltjes

Stieltjes [78]

cf. Zhang [89]

31 Rogel

Rogel [65] and [66]

32 Cesàro

Cesàro [15]

Cesàro [16]

Cesàro [17] and [18, pp. 181–184]

Bromwich [6, p. 201, Chapter VIII, Example B, 35]

33 de la Vallée-Poussin

de la Vallée-Poussin [24]

34 Torelli

Torelli [83]

35 Fibonacci numbers

Landau [54]

36 Knopp

Knopp [49]

37 Generating functions

Hardy and Wright [40, p. 258, Theorem 307]:

Theorem 1.

For f(s)=n=1ann-s and g(s)=n=1bnn-s,


if and only if there is some σ such that


For f(s)=n=1μ(n)n-s and g(s)=1, using [40, p. 250, Theorem 287]


we get

n=1μ(n)xn1-xn=x. (7)

For f(s)=n=1ϕ(n)n-s and


using [40, p. 250, Theorem 288]


we get


For n=p1a1prar, define Ω(n)=a1++an and


For f(s)=n=1λ(n)n-s and


using [40, p. 255, Theorem 300]


we get


We define the von Mangoldt function Λ: by Λ(n)=logp if n is some positive integer power of a prime p, and Λ(n)=0 otherwise. For example, Λ(1)=0, Λ(12)=0, Λ(125)=log5. It is a fact [40, p. 254, Theorem 296] that for any n, the von Mangoldt function satisfies

m|nΛ(m)=logn. (8)

For f(s)=n=1Λ(n)n-s and


using [40, p. 253, Theorem 294]


we obtain


38 Mertens

For Res>1, we define


We also define


Mertens [61] proves the following.

Theorem 2.

As ϱ0,


As ϱ0,


Taking the logarithm,

logζ(1+ϱ)=log(1ϱ)+log(1+γϱ+O(ϱ2))=log(1ϱ)+γϱ+O(ϱ2). (9)

On the other hand, for ϱ>0,


and taking the logarithm,

logζ(1+ϱ) =-plog(1-1p1+ϱ)

Then as ϱ0,


Combining this with (9) we get that as ϱ0,


Mertens [61] also proves that for any x there is some


such that




Mertens shows that


This can be derived using (7), and we do this now; see [58].

Lemma 3.

For Res>1,


For p prime and Res>0,

pdtt(ts-1) =pt-s-111-t-s𝑑t



On the one hand,


On the other hand, for Res>1 we have


Combining these, for Res>1,


Theorem 4.

For any prime p and for m1,


and using this we have

H =m=2p1mpm

Rearranging (7),


With x=t-1,




Thus we have


Using Lemma 3 for s=2,3,4,,


completing the proof. ∎

39 Preliminaries on prime numbers

We define




One sees that


As well,

ψ(x)=m=1px1/mlogp=m=1ϑ(x1/m); (10)

there are only finitely many terms on the right-hand side, as ϑ(x1/m)=0 if x<2m.

Theorem 5.

For x2, ϑ(x)<xlogx, giving

2mlogxlog2ϑ(x1/m) <2mlogxlog2x1/m1mlogx

Thus, using (10) we have


We prove that if limxϑ(x)x=1 then π(x)x/logx=1.

Theorem 6.
lim infxπ(x)x/logx=lim infxϑ(x)x


lim supxπ(x)x/logx=lim supxϑ(x)x.

From (10), ϑ(x)ψ(x). And,





lim infxϑ(x)xlim infxπ(x)x/logx


lim supxϑ(x)xlim supxπ(x)x/logx.

Let 0<α<1. For x>1,


As π(xα)<xα,




This yields

lim infxϑ(x)xαlim infxπ(x)logxx-αlim infxlogxx1-α=αlim infxπ(x)logxx


lim supxϑ(x)xαlim supxπ(x)logxx-αlim supxlogxx1-α=αlim supxπ(x)logxx.

Since these are true for all 0<α<1, we obtain respectively

lim infxϑ(x)xlim infxπ(x)logxx


lim supxϑ(x)xlim supxπ(x)logxx.

40 Wiener’s tauberian theorem

Wiener [85, Chapter III].

Wiener-Ikehara [19]

Rudin [67, p. 229, Theorem 9.7]

We say that a function s:(0,) is slowly decreasing if

lim inf(s(ρv)-s(v))0,v,ρ1+.

Widder [84, p. 211, Theorem 10b]: Wiener’s tauberian theorem tells us that if aL(0,) and is slowly decreasing and if gL1(0,) satisfies




implies that


It is straightforward to check the following by rearranging summation.

Lemma 7.

If n=1anzn has radius of convergence 1, then for |z|<1,


Using Lemma 7 with an=Λ(n) and z=e-x and applying (8), we get

n=1Λ(n)zn1-zn=n=1log(n)zn. (11)

From (11), and Lemma 7 with an=1, we have


We follow Widder [84, p. 231, Theorem 16.6].

Theorem 8.

As x0+,



(1-z)n=1znm=1nam =(1-z)m=1amn=mzn

Using this with am=logm-d(m) and z=e-x gives

n=1(logn-d(n))e-nx =(1-e-x)n=1e-nx(m=1nlogm-m=1nd(m))





we get




One proves that there is some K such that for all 0y<1,


whence, with y=e-x,




and thus we have

n=1(logn-d(n))e-nx =-2γe-x1-e-x+O(x-1/2)











First we show that h is slowly decreasing.

Lemma 9.

h(x) is slowly decreasing.




we have, for 0<x< and ρ>1,

h(ρx)-h(x) =x<nρxΛ(n)-1n

Hence as x and ρ1+,


which shows that h is slowly decreasing. ∎

The following is from Widder [84, pp. 231–232].

Lemma 10.

As x,


Let I(t)=0 for t<0 and I(t)=1 for t0. Writing


we check that for x>0,

0te-xt1-e-xt𝑑h(t) =n=10te-xt1-e-xtΛ(n)-1nd(I(t-n))

On the other hand, integrating by parts,

f(x) =0te-xt1-e-xt𝑑h(t)

By Theorem 8, as x0+,


i.e., as x0+,


Thus, as x,


The following is from Widder [84, p. 232].

Lemma 11.
0t-ixg(t)𝑑t =0t-ixddt(te-t1-e-t)𝑑t



this becomes


If x=0, then using


we get


If x>0, then


By Wiener’s tauberian theorem, it follows that

Lemma 12.

Let I(t)=0 for t<0 and I(t)=1 for t0. Writing


we have

12xd(ψ(t)-[t])t =12x1td(n=1I(t-n)(Λ(n)-1))

Thus, we have established that


41 Hermite

Hermite [42]

Hermite [43]

42 Gerhardt

Gerhardt [33, p. 196] refers to Lambert’s Architectonic.

43 Levi-Civita

Levi-Civita [57]

44 Franel

Franel [32] and [31]

The next theorem shows that the set of points on the unit circle that are singularities of n=1zn1-zn is dense in the unit circle. Titchmarsh [82, pp. 160–161, §4.71].

Theorem 13.

For |z|<1, define


Suppose that p>0,q>1 are relatively prime integers. As r1-,


Set z=re2πip/q and write


On the one hand,

(1-r)n0(modq)zn1-zn =(1-r)m=1zmq1-zmq

as r1.

On the other hand, for n0(modq) we have

|1-zn|2 =|1-rne2πipn/q|2

So far we have not used the hypothesis that n0(modq). We use it to obtain


With this we have


and therefore, as r<1,

(1-r)|n0(modq)zn1-zn| (1-r)n0(modq)|z|n|1-zn|

45 Wigert

The following result is proved by Wigert [86]. Our proof follows Titchmarsh [81, p. 163, Theorem 7.15]. Cf. Landau [55].

Theorem 14.

For λ<12π and N1,


as z0 in any angle |argz|λ.


For σ>1, s=σ+it,


Using this, for Rez>0 we have

12πi2-i2+iΓ(s)ζ2(s)z-s𝑑s =n=1d(n)12πi2-i2+iΓ(s)(nz)-s𝑑s
=n=1d(n)e-nz. (12)

Define F(s)=Γ(s)ζs(s)z-s. F has poles at 1,0, and the negative odd integers. (At each negative even integer, Γ has a first order pole but ζ2 has a second order zero.) First we determine the residue of F at 1. We use the asymptotic formula


the asymptotic formula


and the asymptotic formula


to obtain

Γ(s)ζs(s)z-s =(1-γ(s-1)+O(|s-1|2))(1(s-1)2+2γs-1+O(|s-1|2))

Hence the residue of F at 1 is


Now we determine the residue of F at 0. The residue of Γ at 0 is 1, and hence the residue of F at 0 is


Finally, for n0 we determine the residue of F at -(2n+1). The residue of Γ at -(2n+1) is (-1)2n+1(2n+1)!, hence the residue of F at -(2n+1) is




Let M>0, and let C be the rectangular path starting at 2-iM, then going to 2+iM, then going to -2N+iM, then going to -2N-iM, and then ending at 2-iM. By the residue theorem,

CF(s)𝑑s=2πi(γz-logzz+14+n=0N-1-B2n+22(2n+2)!(2n+2)z2n+1). (13)

Denote the right-hand sideof (13) by 2πiR. We have


We shall show that the second and fourth integrals tend to 0 as M. For s=σ+it with -2Nσ2, Stirling’s formula [82, p. 151] tells us that


As well [81, p. 95], there is some K>0 such that in the half-plane σ-2N,



z-s =e-slogz

and so for |argz|λ,




and because λ<π2 this tends to 0 as M. Likewise,


as M. It follows that




We bound the integral on the right-hand side. We have


The first integral satisfies


because Γ(s)ζ2(s) is continuous on the path of integration. The second integral satisfies

|σ=-2N,|t|>1F(s)𝑑s| σ=-2N,|t|>1e-π2|t||t|σ-12|t|K|z|-σeλ|t|𝑑s

because λ<π2. This establishes


Using (12) and (13), this becomes


completing the proof. ∎

For example, as B2=16,B4=-130,B6=142, the above theorem tells us that


46 Steffensen

Steffensen [75]

47 Szegő

Szegő [79]

48 Pólya and Szegő

Pólya and Szegő [64]

49 Partition function



Taking the logarithm,


and switching the order of summation gives


On the one hand, for 0<x<1 we have mxm-1(1-x)<1-xm and using this,


On the other hand, for -1<x<1 we have 1-xm<m(1-x), and using this, for 0<x<1 we have


Thus, for 0<x<1,


Taking x1- gives




See Stein and Shakarchi [76, p. 311].

50 Hansen

Hansen [39]

51 Kiseljak

Kiseljak [45]

52 Unsorted

In 1892, in volume VII, no. 23, p. 296 of the weekly Naturwissenschaftliche Rundschau, it is stated that for the year 1893, one of the six prize questions for the Belgian Academy of Sciences in Brussels is to determine the sum of the Lambert series


or if one cannot do this, to find a differential equation that determines the function.

Gram [35] on distribution of prime numbers.

Hardy [41]

Bohr and Cramer [1, p. 820]

Flajolet, Gourdon and Dumas [30]


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