Rapidly Converging Formulae for $\Zeta (4K\Pm 1) $

$Rapidly Converging Formulae for $\Zeta (4K\Pm 1) $$

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2. Lambert series In this section we focus on the Lambert series at negative odd integer arguments and x = 1. Since x is set to 1 in this paper, to avoid redundancy we use the notation Lq(s, 1) Lq(s). ≡ Theorem 2.1. For odd negative integers s the Lambert series at x = 1 satisﬁes the following relations: (1) For s = 1, − 1 π Le 2πt ( 1) L 2π ( 1) = log t sinh(log t). − − − e− t − 2 − 6 (2) For s = (4k 1) = 3, 7, 11,..., − − − − −

1 2k 1 Le 2πt ( 4k +1)+ t − L 2π ( 4k + 1) 2k 1 − e− t t − − − k j+1 4k 1 ( 1) B2j B4k 2j cosh[(2k 2j) log t] = (2π) − − − − (2j)! (4k 2j)! (1 + δjk) j=0 X − ζ(4k 1) cosh[(2k 1) log t]. − − − (3) For s = (4k +1)= 5, 9, 13,..., − − − −

1 2k Le 2πt ( 4k 1) t L 2π ( 4k 1) t2k − − − − e− t − − k j+1 4k+1 ( 1) B2j B4k+2 2j sinh[(2k +1 2j) log t] = (2π) − − − (2j)! (4k +2 2j)! j=0 X − + ζ(4k + 1) sinh(2k log t),

where Bk is the kth Bernoulli number and δjk is the Kronecker delta.

Remark Although the theorem is stated for negative odd integer arguments of the Lambert series, it holds true for positive odd arguments as well. The results for positive odd arguments, obtained by using negative k values in the theorem, reproduce modular properties of the Eisenstein series (1.3).

Proof. For the s = 1 case, by taking the logarithm of the complete expansion of the q-Pochhammer− symbol provided in the Referee Remark 3.3 of Ref. [3] we get

2 π 1 t t ∞ 4nπ2 Lq( 1) = + log + Li e− t − 6t 2 2π − 24 1 (2.1) n=1 X π2 1 t t = + log + L 4π2 ( 1), 6t 2 2π − 24 e− t −

t where q = e− and Lis(q) is the polylogarithm function. Replacing t with 2πt and rearranging the terms completes the proof. For s = 4k+1 we begin with the result in Corollary 2.4(2) of Ref. [3]. Replacing − t with 2πt, writing ζ ′(2 4k) in terms of ζ(4k 1), and writing the zeta function − − RAPIDLY CONVERGING FORMULAE FOR ζ(4k 1) 3 ± at even arguments in terms of Bernoulli numbers gives 1 L 2πt ( 4k + 1) 2k 1 e− t − − k 1 k+1 2 − j+1 4k 1 ( 1) B2k ( 1) B2j B4k 2j cosh[(2k 2j) log t] (2π) − − + − − − ≈  (2k)!2 2 (2j)! (4k 2j)!  j=0 X − ζ(4k 1) cosh[(2k 1) log t].  − − − The right hand side of the equation above is completely symmetric with respect to 2k 1 a t 1/t transformation. Thus adding t − L 2π ( 4k + 1) to the left hand side → e− t − of the equation to make it completely symmetric yields the result in the theorem. Note that such symmetry arguments only guarantee numerical accuracy. However, a comparison with Entry 21(i) in [4] shows that the result is exact, and there are 2 no other error terms. In the ﬁnal result the B2k term is included in the sum over j by using δjk and extending the limit of the sum from k 1 to k. Similarly, for s = 4k 1 case from Corollary 2.4(2) of− Ref. [3] we obtain − − 1 L 2πt ( 4k 1) t2k e− − − k j+1 4k+1 ( 1) B2j B4k+2 2j sinh[(2k +1 2j) log t] (2π) − − − ≈ (2j)! (4k +2 2j)! j=0 X − + ζ(4k + 1) sinh(2k log t). In this case the right hand side of the equation is antisymmetric with respect to a 2k t 1/t transformation. Thus, we add t L 2π ( 4k 1) to the left hand side of → e− t − − the equation to make the equation completely antisymmetric. As in the case above, a comparison with Entry 21(i) in [4] shows that the result is exact, which yields the result in the theorem.

Corollary 2.2. For positive odd integers (4k 1)=3, 7, 11, ... we have −

k j+1 4k 1 ( 1) B2j B4k 2j − L 2π ζ(4k 1) = (2π) − − 2 e− ( 4k + 1). − (2j)! (4k 2j)! (1 + δjk) − − j=0 X − Proof. Substituting t = 1 in Theorem 2.1(2) and solving for ζ(4k 1) completes the proof. −

Corollary 2.3. For positive odd integers (4k +1)=5, 9, 13, ... we have

k j 4k+1 ( 1) (2k +1 2j)B2jB4k+2 2j ζ(4k +1)= (2π) − − − 2k (2j)! (4k +2 2j)! j=0 X − 2πL 2πe− e′ 2π ( 4k 1) − − − 2L 2π ( 4k 1) , − k − e− − − where L′ = dL/dq. Proof. Diﬀerentiating Theorem 2.1(3) with respect to t at t = 1 and solving for ζ(4k + 1) completes the proof. 4 SHUBHOBANERJEEANDBLAKEWILKERSON

Remark The results in Corollaries 2.2 and 2.3 are well known [4–6] but have been reproduced here for the sake of completeness. The results hold true for negative values of k as well.

3. Formulae for ζ(4k + 1) In Corollary 2.3 we established a formula for ζ(4k +1) that involved the Lambert 2π series and its derivative with respect to q at q = e− . In this section we establish three new formulae for ζ(4k +1) that are in terms of the Lambert series only, and not its derivative. First, we prove a general identity for the Lambert series that is true for all s values in the form of the following lemma.

Lemma 3.1. For any positive prime integer p and s C the Lambert series at x =1 satisﬁes: ∈

p 1 − L s+1 L s+1L 1 i 2πn (s) = (p + p) q(s) p qp (s). q p e p − n=0 X Proof. Writing the Lambert series in terms of the divisor function using (1.2) gives

p 1 p 1 1 − 1 − ∞ k i 2πnk L p p 1 i 2πn (s)= σs(k)q e p q p e p p n=0 n=0 k X X X=1 p 1 1 ∞ k − i 2πnk = σ (k)q p e p p s k n=0 X=1 X ∞ l = σs(lp) q l X=1 s ∞ l s ∞ l l = (p + 1) σs(l) q p σs( ) q − p l=1 p l X X| s ∞ l s ∞ mp = (p + 1) σs(l) q p σs(m) q , − l m=1 X=1 X where p l indicates sum only over integers l that are divisible by p. Rewriting the two inﬁnite| sums in terms of Lambert series using (1.2) and multiplying by p on both sides proves the lemma. In the proof above we used the following property of the divisor function [1]

s s l σs(lp) = (p + 1)σs(l) p σs( ), − p l where σs( p ) is non-zero only when p divides l.

Remark For p = 2 the limits of the sum over n in the lemma can be replaced with the symmetric limits6 ranging from (p 1)/2 to (p 1)/2. It is this symmetric range that we use in later proofs when− p−= 3 and p =− 5. RAPIDLY CONVERGING FORMULAE FOR ζ(4k 1) 5 ± Theorem 3.2. The Riemann zeta function at odd integers 4k +1 other than unity 2π 4π can be written in terms of Lambert series at q = e− and e− as:

k j+1 4k+1 ( 1) bjk B2j B4k+2 2j ζ(4k +1)= (2π) − − (2j)!(4k +2 2j)! j=0 X − 2ak +4 4 L 2π L 4π e− ( 4k 1)+ e− ( 4k 1), − ak − − ak − − where 4k+1 k 2k ak =2 ( 1) 2 1, − − − 2j 1 4k+1 2j 4k+1 2j 2j 1 2 − 1+(1+ i) − 2 − 1+(1+ i) − bjk = − . ak 1 Proof. Applying Theorem 2.1(3) at t = 2 gives

2k 1 2 L π ( 4k 1) L 4π ( 4k 1) e− − − − 22k e− − − k j+1 2k+1 2j 4k+1 ( 1) B2j B4k+2 2j sinh(log 2 − ) (3.1) = (2π) − − − (2j)! (4k +2 2j)! j=0 X − ζ(4k + 1) sinh(log 22k). − Applying Theorem 2.1(3) at t = 1 i =1/(1 + i) gives 2 − 2 2k 1 (1 + i) L e π ( 4k 1, 1) Le 2π ( 4k 1) − − − − − (1 + i)2k − − − k j+1 2k+1 2j (3.2) 4k+1 ( 1) B2j B4k+2 2j sinh[log(1 + i) − ] = (2π) − − − (2j)! (4k +2 2j)! j=0 X − ζ(4k + 1) sinh[log(1 + i)2k]. − 2π Applying Lemma 3.1 at q = e− with p = 2 gives

L e π ( 4k 1)+ Le π ( 4k 1) − − − − − − − (3.3) 4k 4k = (2− + 2)L 2π ( 4k 1) 2− L 4π ( 4k 1). e− − − − e− − −

Eliminating L e π ( 4k 1) and Le π ( 4k 1) from Equations (3.1), (3.2), and − − − − − − − L 2π L 4π (3.3) and solving for ζ(4k +1) in terms of e− ( 4k 1) and e− ( 4k 1) completes the proof of the theorem. − − − −

Theorem 3.3. The Riemann zeta function at odd integers 4k +1 other than unity 3π 4π 6π can be written in terms of Lambert series at q = e− , e− , and e− as: − 4k+1 k j+1 k 2k+1 (2π) ( 1) cjk B2j B4k+2 2j ( 1) 2 − L 3π ζ(4k +1)= − + − e− ( 4k 1) 2bk (2j)!(4k +2 2j)! bk − − − j=0 X − 2k ak 2 L 4π ( 4k 1) + L 6π ( 4k 1) 4k 1 e− 2k 1 e− − 2 − bk − − 2 − bk − − 6 SHUBHOBANERJEEANDBLAKEWILKERSON where 24k 34k+1 +1 ak = 4k+1 k 2k , 2 ( 1) 2 +1 4k+1 − − 4k+1 k 2k 3 1 k 2k 2 ( 1) 2 1 bk = − ( 1) 2 ak − − − , 2 − − − 24k+1 1 1 2j 1 4k+1 2j 4k+1 2j 2j 1 cjk =3 − (1 + in) − 3 − (1 + in) − − n= 1 n= 1 X− X− 4k+1 2j 2j 1 1+(1+ i) − 1+(1+ i) − ak . − 24k+1 2j − 22j 1 − − Remark This formula converges at about 4.09 digits for each additional term in the Lambert series. The ﬁrst few examples of this theorem are listed in the Appendix. 3π If needed, the Lambert series at q = e− can be replaced by Lambert series at − 6π positive only q values using Lemma 3.1 at q = e− with p = 2,

4k L e 3π ( 4k 1) = Le 3π ( 4k 1)+(2− + 2)Le 6π ( 4k 1) − − − − − − − − − − − 4k 2− L 12π ( 4k 1). − e− − − However, the result as stated in the theorem is more compact and has the same rate of convergence as the one with positive q values. Similar replacement for Lambert series at negative q values can be made in Theorems 3.4 and 4.3 as well.

Proof. The proof is a generalization of the proof of Theorem3.2 to the p = 3 case. Applying Theorem 2.1(3) at t = 1 and t = 1 i yields three equations involving 3 3 ± 3 L 2π ( 4k 1), L 2π 2πi ( 4k 1), and ζ(4k +1) in terms of L e 3π ( 4k 1), e− 3 e− 3 ± 3 − − − − − − 2π − − and L 6π ( 4k 1). Applying Lemma 3.1 at q = e− with p = 3 gives e− − − 1 L 4k L 4kL 2π +i 2πn ( 4k 1) = (3− + 3) e 2π ( 4k 1) 3− e 6π ( 4k 1). e− 3 3 − − − − − − − − − n= 1 X−

L 2π L 4π Theorem 3.2 supplies the ﬁfth equation that relates e− ( 4k 1) to e− ( 4k 1) and ζ(4k + 1). − − − − Using these ﬁve equations to eliminate L 2π ( 4k 1), L 2π 2πi ( 4k 1), and e− 3 − − e− 3 ± 3 − − Le 2π ( 4k 1) and solving for ζ(4k+1) in terms of L e 3π ( 4k 1), Le 4π ( 4k − − − − − − − − − − 1), and L 6π ( 4k 1) completes the proof of the theorem. e− − −

Theorem 3.4. The Riemann zeta function at odd integers 4k +1 other than unity 4π 5π 10π can be written in terms of Lambert series at q = e− , e− , and e− as: −

4k+1 k j+1 (2π) ( 1) cjk B2j B4k+2 2j 1 ζ(4k +1)= − L 4π ( 4k 1) − 4k 1 e− 2bk (2j)!(4k +2 2j)! − 2 bk − − j=0 − X − k 2k+1 ( 1) 2 ak 2ak L 5π L 10π + − e− ( 4k 1)+ e− ( 4k 1), bk − − − bk − − RAPIDLY CONVERGING FORMULAE FOR ζ(4k 1) 7 ± where 24k+1 ( 1)k22k +1 a = , k 4k 4k+1 − 2−k 1 2 5 2 5 cos(4k tan− 2)+1 − × 2 4k+1 k 2k ak 4k+1 4k 2 ( 1) 2 1 bk = 5 (1 + in) − − − , 2 − − 24k+1 n= 2 X− 2 2 2j 1 4k+1 2j 4k+1 2j 2j 1 cjk = ak 5 − (1 + in) − 5 − (1 + in) − − n= 2 n= 2 X− X− 1+(1+ i)4k+1 2j 1+(1+ i)2j 1 − − . − 24k+1 2j − 22j 1 − − Remark This formula has a convergence rate of about 5.45 digits per term and is our best result for ζ(4k + 1). The ﬁrst few examples of this theorem are listed in the Appendix.

Proof. The proof of this theorem is very similar to that of Theorem 3.3. Theo- 1 1 i 1 2i rem 2.1(3) is now applied at t = 5 , 5 5 , and 5 5 and Lemma 3.1 is applied at 2π ± ± q = e− with p = 5 instead of p = 3. Rest of the logic remains the same.

4. Additional formulae for ζ(4k + 1) In the previous section we provided formulae for ζ(4k + 1) in terms of Lambert 2npπ series at q values of the form e− where p was a prime equal to 2, 3, or 5 and n was 0, 1, or 2. In this section we provide an additional class of formulae for ζ(4k+1) 2n√mπ using Lambert series with q values of the form e− where m is equal to 3, 7, or 15 and n is 0, 1, or 2. First we state the following two lemmas on which the theorems in this section are based. Both lemmas combine two Lambert series to produce a series with a stronger convergence rate per term. Lemma 4.1. For any s C the Lambert series at x =1 satisﬁes: ∈ L L s+1 L 2s+2 s+1 L i√q(s)+ i√q(s)= (2 + 2) q(s)+(2 +3 2 + 4) q2 (s) − − × 2s+2 s+2 (2 +2 ) L 4 (s). − q Proof. Applying Lemma 3.1 at p = 2 and replacing q with q gives − L L s+1 L s+1L (4.1) i√q(s)+ i√q(s)=(2 + 2) q(s) 2 q2 (s). − − − Now using Lemma 3.1 at p = 2 and replacing q with q2 gives s+1 s+1 (4.2) Lq(s)+ L q(s)=(2 + 2)Lq2 (s) 2 Lq4 (s). − − Solving for L q(s) in the second equation and replacing it in the ﬁrst equation above completes− the proof.

Lemma 4.2. For any s C the Lambert series at x =1 satisﬁes: ∈ ∞ L L n+1 s 1 q(s) i i√q(s) i√q(s) = ( 1) (2n + 1) sech (n + ) log q . S ≡ − − − 2 n=0 X 8 SHUBHOBANERJEEANDBLAKEWILKERSON

Proof. ∞ ns(iq)n ∞ ns( iq)n Liq(s) L iq(s)= − − − 1 (iq)n − 1 ( iq)n n=1 n=1 X − X − − ∞ (2m + 1)s(iq)2m+1 ∞ (2m + 1)s( iq)2m+1 = − 1 (iq)2m+1 − 1 ( iq)2m+1 m=0 m=0 X − X − − ∞ 2(2m + 1)s(iq)2m+1 = 1 (iq)4m+2 m=0 X − ∞ ( 1)m+1(2m + 1)sq2m+1 = 2i − − 1+ q4m+2 m=0 X ∞ = i ( 1)m+1(2m + 1)ssech[(2m + 1) log q] . − − m=0 X Replacing q with √q and multiplying by i on both sides completes the proof. Now we state the three formulae for ζ(4k + 1) in increasing order of their convergence rates in the form of the following theorems. Theorem 4.3. The Riemann zeta function at odd integers 4k +1 other than unity √3π can be written in terms of the Lambert series at q = e− as: − k j 4k+1 ( 1) bjk B2j B4k+2 2j L ζ(4k +1) =(2π) − − ak e √3π ( 4k 1) (2j)!(4k +2 2j)! − − − − − j=0 X − where 24k+1 +1 ak = , 24k cos 2πk − 3 2j 1 (2k 1 j)π 4k+1 2j (j+1)π 2 − sin −3− +2 − sin 3 bjk = . 24k cos 2πk − 3 Proof. Substituting t = (√3 i)/2 in Theorem 2.1(3) and adding gives one equation L ± in e √3π ( 4k 1) and ζ(4k +1). Solving for ζ(4k +1) yields the ﬁnal result. − − − − Theorem 4.4. The Riemann zeta function at odd integers 4k +1 other than unity √7π 2√7π 4√7π can be written in terms of the Lambert series at q = e− , e− , and e− as: k j 4k+1 ( 1) cjk B2j B4k+2 2j ζ(4k +1) =(2π) − − + akL √7π ( 4k 1) (2j)!(4k +2 2j)! e− − − j=0 X − 4k + bkL 2√7π ( 4k 1)+2− akL 4√7π ( 4k 1), e− − − e− − − where 4k 2k+1 2+2− 2− cos(4kθ) a = − , k 1 2 2k cos(4kθ) − − 4k 8k 2k+2 6k+1 4+3 2− +2− 2− cos(4kθ) 2− cos(4kθ) bk = × − − , − 1 2 2k cos(4kθ) − − 2k+ 1 j j 1 2 2 − cos[(2j 1)θ] 2 − 2 cos[(4k +1 2j)θ] c = − − − , jk 22k cos(4kθ) − RAPIDLY CONVERGING FORMULAE FOR ζ(4k 1) 9 ±

1 and θ = cot− √7. Proof. Substituting t = (√7 i)/4 in Theorem 2.1(3) and adding gives one equa- ± tion in L √7π/2 ( 4k 1), L √7π ( 4k 1), and ζ(4k + 1). Eliminating ie− − − e− − − ± L − √7π the sum of √7π/2 ( 4k 1) using Lemma 4.1 at q = e− , and eliminat- ie− − − L ± 2√7π ing √7π ( 4k 1) using Lemma 3.1 for p = 2 at q = e− yields the ﬁnal e− result.− − −

Theorem 4.5. The Riemann zeta function at odd integers 4k +1 other than unity √15π 2√15π 4√15π can be written in terms of the Lambert series at q = e− , e− , and e− , and the series √15π ( 4k 1) as: Se− − − k j 4k+1 ( 1) cjk B2j B4k+2 2j ζ(4k +1) =(2π) − − + cot(2kθ) √15π ( 4k 1) (2j)!(4k +2 2j)! Se− − − j=0 X − 4k + 2− +2 L √15π ( 4k 1) e− − − 8k 4k L 2− +3 2− +4 2√15π ( 4k 1) − × e− − − 8k 4k+1 L + 2− +2− 4√15π ( 4k 1), e− − − where

cjk = csc(2kθ) sin[(2k+1 2j) θ], − 1 and θ = cot− √15. Remark This result converges almost as rapidly as Theorem 3.4 at 5.28 decimal digits per term. Some examples of this theorem are listed in the Appendix.

Proof. Substituting t = (√15 i)/4 in Theorem 2.1(3) and adding gives one equa- ± tion containing both the sum and difference of L √15π/2 ( 4k 1), and ζ(4k +1). ie− − − L ± √15π Eliminating the sum of √15π/2 ( 4k 1) using Lemma 4.1 at q = e− , and ie− − − ± L √15π eliminating the difference of √15π/2 ( 4k 1) using Lemma 4.2 at q = e− ie− yields the final result. ± − −

5. Formulae for ζ(4k 1) − The methods of Section 4 are mirrored with minor variations to produce similar results for ζ(4k 1) stated in terms of the three theorems below. − Theorem 5.1. The Riemann zeta function at odd integers 4k 1 can be written √3π 2√3π 4√3π − in terms of Lambert series at q = e− , e− , and e− as:

k j+1 4k+1 4k 1 ( 1) bjk B2j B4k 2j 2 +4L ζ(4k 1) =(2π) − − − e √3π ( 4k + 1) − (2j)!(4k 2j)! − ak − − j=0 X − 4k+2 4k+4 (2k 1)π 2 +2− +12+8cos −3 L 2√3π ( 4k + 1) e− − ak − 4k+4 2− +8L + e 4√3π ( 4k + 1), ak − − 10 SHUBHOBANERJEEANDBLAKEWILKERSON where 4k (4k+1)π ak =2 + 4 sin 6 , 4k+1 2j (2j 1)π 2j+1 (4k 1 2j)π 2 − cos −6 +2 cos − 6− bjk = . ak(1 + δjk) Proof. Substituting t = (√3 i)/4 in Theorem 2.1(2) and adding the results gives L ± L one equation in ie √3π/2 ( 4k+1), e 2√3π ( 4k+1), and ζ(4k 1). Eliminating ± − − − − − eliminating L √3π/2 ( 4k + 1) in favor of L √3π ( 4k + 1), L 2√3π ( 4k + 1), ie− − e− − e− − L ± − 2√3π and 4√3π ( 4k + 1) using Lemma 3.1 for p = 2 at q = e− yields the ﬁnal e− result. − Theorem 5.2. The Riemann zeta function at odd integers 4k 1 can be written √7π 2√7π 4√7π − in terms of Lambert series at q = e− , e− , and e− as: k j+1 4k 1 ( 1) cjk B2j B4k 2j bk L ζ(4k 1) =(2π) − − − + e √7π ( 4k + 1) − (2j)!(4k 2j)! ak − − j=0 X − 4k+2 2k+2 (2− + 2) bk 2− L − e 2√7π ( 4k + 1) − ak − − 4k+2 2− bk L + e 4√7π ( 4k + 1), ak − − where 2k ak =2 + 2 cos[(4k 2) θ], − 2k+1 2k+2 3 bk =2 +2− +2 2 sin[(4k+1) θ]+ 2cos(4kθ), 2k+ 1 j j+ 1 2 2 − cos[(2j 1) θ]+2 2 cos[(4k 1 2j) θ] cjk = − − − , ak(1 + δjk) 1 and θ = cot− √7. Remark This formula for ζ(4k 1) has a convergence rate of about 3.6 decimal digits per term which is about− 0.9 decimal digits per term faster than that of Corollary 2.2. Some examples of this theorem are listed in the Appendix. Proof. Substituting t = (√7 i)/4 in Theorem 2.1(2) and adding the results gives ± one equation in L √7π/2 ( 4k+1), L √7π ( 4k+1), and ζ(4k 1). Eliminating ie− − e− − − L ± − √7π the sum of √7π/2 ( 4k + 1) using Lemma 4.1 at q = e− , and eliminating ie− − L ± 2√7π √7π ( 4k + 1) using Lemma 3.1 for p = 2 at q = e− yields the ﬁnal e− result.− − Theorem 5.3. The Riemann zeta function at odd integers 4k 1 can be written √15π 2√15π 4√15−π in terms of Lambert series at q = e− , e− , and e− , and the series

√15π ( 4k + 1) as: Se− − k j+1 4k 1 ( 1) cjk B2j B4k 2j ζ(4k 1) =(2π) − − − + ak √15π ( 4k + 1) − (2j)!(4k 2j)! Se− − j=0 X − 4k+2 + 2− +2 bkL √15π ( 4k + 1) e− − 8k+4 4k+2 L 2− +3 2− +4 bk 2√15π ( 4k + 1) − × e− − 8k+4 4k+3 L + 2− +2− bk 4√15π ( 4k + 1), e− − RAPIDLY CONVERGING FORMULAE FOR ζ(4k 1) 11 ± where cos[(4k 1)θ] 2 sin(4kθ) a = − − , k 4cos2[(2k 1)θ] − 2 + 2cos(4kθ) + sin[(4k 1)θ] b = − , k 4cos2[(2k 1)θ] − cos[(2k 2j)θ] cjk = − , (1 + δjk) cos[(2k 1)θ] − 1 and θ = cot− √15. Remark This formula has a convergence rate of about 5.28 decimal digits per term which is the best of our results for ζ(4k 1). Some examples of this theorem are listed in the Appendix. − Proof. Substituting t = (√15 i)/4 in Theorem 2.1(2) and adding the results gives ± L one equation containing both the sum and difference of ie √15π/2 ( 4k + 1), ± − − and ζ(4k 1). Eliminating the sum of L √15π/2 ( 4k + 1) using Lemma 4.1 at − ie− − √15π ± L q = e− , and eliminating the difference of √15π/2 ( 4k+1) using Lemma 4.2 ie− − √15π ± at q = e− yields the final result.

6. Additional results for the Lambert series Some other interesting results can be arrived at through the manipulation of results in the earlier sections. For example, from Theorem 2.1 we get

Theorem 6.1. For any constant a with Re(a) > 0, the Lambert series Lq(s) at negative odd integers s satisﬁes the following: (1) For s = (4k + 1) = -1, -5, -9, ..., − 1 2k 2k 2k a L 2πt ( 4k 1) (a + a− )Le 2πt ( 4k 1) t2k e− a − − − − − − h 2k + a− L 2πat ( 4k 1) e− − − 2k 2k 2ik 2k t a L 2π ( 4k 1) (a + a− )L 2π ( 4k 1) − e− at − − − e− t − − h 2k + a− L 2πa ( 4k 1) e− t − − k j+1 i 4k+1 ( 1) bjk(a)B2j B4k+2 2j sinh[(2k +1 2j) log t] = (2π) − − − . (2j)! (4k +2 2j)! j=0 X − (2) For s = (4k 1) = -3, -7, -11, . . . , − − 1 2k 1 2k 1 2k+1 a − L 2πt ( 4k + 1) (a − + a− )Le 2πt ( 4k + 1) 2k 1 e− a − t − − − − h 2k+1 + a− L 2πat ( 4k + 1) e− − 2k 1 2k 1 i2k 1 2k+1 + t − a − L 2π ( 4k + 1) (a − + a− )L 2π ( 4k + 1) e− at − − e− t − h 2k+1 + a− L 2πa ( 4k + 1) e− t − k j i 4k 1 ( 1) cjk(a)B2j B4k 2j cosh[(2k 2j) log t] = (2π) − − − − , (2j)! (4k 2j)! j=0 X − 12 SHUBHOBANERJEEANDBLAKEWILKERSON

where 2k 2k 2k+1 2j 2k 1+2j bjk(a)= a + a− a − a− − , − − and 2k 1 2k+1 2k 2j 2k+2j a − + a− a − a− cjk(a)= − − . (1 + δjk) Proof. The proof follows from making the substitutions t at and t t/a in Theorem 2.1 and then combining them with the original result→ in Theore→m 2.1 in the manner stated in the theorem above. The term containing the zeta function at odd arguments, ζ(4k 1), is eliminated. The special case of s = 1 is included as the k = 0 case in s =± (4k + 1). The log t term is eliminated in this− case. − As examples of Theorem6.1, odd powers of π can be calculated in terms of the Lambert series at desired q values as stated below. 1 i 1 Example By choosing t = 2 2 and a = 2 in Theorem 6.1(1), eliminating ± π L e π ( 4k 1) using Lemma 3.1 at q = e− and p = 2, and eliminating the − − − − sum L π ( 4k 1)+ L π ( 4k 1) using Lemma 4.1 we get ie− 2 ie− 2 − − − − −

π = 72 L π ( 1) 96 L 2π ( 1)+24 L 4π ( 1), e− − − e− − e− − 5 π = 7056 L π ( 5) 6993 L 2π ( 5) 63 L 4π ( 5), e− − − e− − − e− − 9 28226880 112920885 13365 π = L π ( 9) L 2π ( 9) + L 4π ( 9). 41 e− − − 164 e− − 164 e− − 1 Example By choosing t = 1 and a = 2 in Theorem 6.1(2) we get

3 π = 720 L π ( 3) 900 L 2π ( 3)+ 180 L 4π ( 3), e− − − e− − e− − 7 907200 14175 π = L π ( 7) 70875 L 2π ( 7)+ L 4π ( 7), 13 e− − − e− − 13 e− − 11 27243216000 218158565625 212837625 π = L π( 11) L 2π( 11)+ L 4π( 11). 4009 e− − − 32072 e− − 32072 e− − Remark The results for the odd powers of π in the two examples above agree with those listed by Plouﬀe [7].

Proposition 6.2. The results of Theorems 3.3 and 3.4 can be combined to produce faster converging series for π4k+1 (than in the example above) in terms of the 3π 4π 5π 6π 10π Lambert series at q = e− , e− , e− , e− , and e− . For example, − − 5 π = 3686634 L e 3π( 5) + 2463048 Le 4π ( 5) + 402570 L e 5π ( 5) − − − − − − − − − 1843317 201285 + L 6π ( 5) L 10π ( 5). 2 e− − − 2 e− − Remark Combining Theorems 3.4 and 4.5 gives the fastest converging, but not the cleanest, result for π4k+1. Similarly,

Proposition 6.3. The results of Corollary 2.2 and Theorem 5.2 can be combined to 4k 1 produce a faster converging series for π − (than in the example above) in terms RAPIDLY CONVERGING FORMULAE FOR ζ(4k 1) 13 ±

2π √7π 2√7π 4√7π of the Lambert series at q = e− , e− , e− , and e− . For example, 3 3960 L 4320 L π = e 2π ( 3)+ √7π ( 3) 77 29√7 − − 77 29√7 e− − − 9360 − 1080 L 2√7π ( 3)+ L 4√7π ( 3). − 77 29√7 e− − 77 29√7 e− − − − Remark Combining the results of Theorems 5.2 and 5.3 gives the fastest converg- 4k 1 ing result for π − . Proposition 6.4. By using the methods outlined in Section 3 but with s = 1 we can calculate the logarithm of the ﬁrst three primes in terms of the Lambert series− : 2π 8 8 log 2 = L 2π ( 1)+ L 4π ( 1), 9 − 3 e− − 3 e− − 19π 32 4 8 16 log 3 = Le 2π ( 1)+ L e 3π ( 1)+ Le 4π ( 1)+ Le 6π ( 1), 54 − 9 − − 3 − − − 9 − − 3 − − 37π 8 2 log 5 = Le 2π ( 1)+ Le 4π ( 1)+ L e 5π ( 1)+ Le 10π ( 1). 72 − 3 − − 3 − − − − − − − Acknowledgments We thank Van Vleet Physics Professorship and Mac Armour Physics Fellowship for their support. We thank Simon Plouﬀe for making the numerically computed results for ζ(4k 1) available online; we found them to be extremely helpful in formulating our± results. We are grateful to Jonathan Sondow and Eric Weisstein for maintaining a comprehensive web page on the Riemann zeta function. References

[1] M. Abramowitz and I. Stegun, Handbook of Mathematical Functions with Formulas, Graphs, and Mathematical Tables, (Dover, New York: Dover, 1972). [2] T. M. Apostol, Modular Functions and Dirichlet Series in Number Theory, 2nd ed. (Springer, New York, 1997). [3] S. Banerjee and B. Wilkerson, Asymptotic expansions of Lambert series and related q-series, Int. J. Number Theory 13 (2017) 2097–2113. [4] B. C. Berndt, Ramanujan’s Notebooks, Part II, Ch. 14. (Springer, New York, 1988). [5] H. Cohen, High Precision Computation of Hardy-Littlewood Constants, (2000). Preprint. http://www.math.u-bordeaux.fr/∼cohen/hardylw.dvi [6] J. Sondow and E. W. Weisstein, “Riemann Zeta Function.” From MathWorld–A Wolfram Web Resource. http://mathworld.wolfram.com/RiemannZetaFunction.html [7] S. Plouﬀe, Identities Inspired from Ramanujan Notebooks (Part 2), (2006). http://www.lacim.uqam.ca/∼plouﬀe/inspired2.pdf 14 SHUBHOBANERJEEANDBLAKEWILKERSON

Appendix We list examples of the first few ζ(4k + 1) generated by Theorem3.3. For com- 4k+1 pactness only the coefficients are listed respectively for π , L e 3π ( 4k 1), − − − − L 4π ( 4k 1), and L 6π ( 4k 1). e− − − e− − − 682 296 488 74 ζ(5) = , , , 201285 −355 −355 355 5048 2272 5624 142 ζ(9) = , , , 150155775 1605 −1605 1605 21462388 1056896 3188648 16514 ζ(13) = , , , 62314387009875 −2114515 −2114515 2114515 12292037116 66978304 258280328 261634 ζ(17) = , , , 3476479836810605625 95520195 − 95520195 95520195 203055579851796692 4297066496 20920706408 ζ(21) = , , , 5594631411704844933908859375 −12606788275 −12606788275 4196354 12606788275 91295430825021344 274844360704 1694577218888 ζ(25) = , , , 245007095801727658882798940625 709832878755 − 709832878755 67100674 709832878755 Examples of the first few ζ(4k+1) generated by Theorem3.4. The coefficients are 4k+1 L 4π L 5π L 10π listed respectively for π , e− ( 4k 1), e− ( 4k 1), and e− ( 4k 1). The result for ζ(5) matches that of− Plouffe− [7] when− converted− − to positive −q values.−

694 6280 296 74 ζ(5) = , , , 204813 −3251 −3251 3251 6118928 3908360 15904 994 ζ(9) = , , , 182032863705 −1945731 1945731 1945731 4131911428 2441359240 1056896 16514 ζ(13) = , , , 11996181573401025 −1221199811 −1221199811 1221199811 687182059214356 1525878246920 66978304 ζ(17) = , , , 194362869568557017703375 − 762905503491 762905503491 261634 762905503491 2560199089127112465412 953674355019400 ζ(21) = , , 70537137132904905751999929343125 −476839323944771 4297066496 4196354 , −476839323944771 476839323944771 114987316346581920808496 596046447625404680 ζ(25) = , , 308598430935986470640664020644801875 −298023086356971651 274844360704 67100674 , 298023086356971651 298023086356971651 RAPIDLY CONVERGING FORMULAE FOR ζ(4k 1) 15 ± Examples of the ﬁrst few ζ(4k+1) generated by Theorem4.4. The coeﬃcients are 4k+1 listed respectively for π , L √7π ( 4k 1), L 2√7π ( 4k 1), L 4√7π ( 4k 1). e− − − e− − − e− − −

5 64 130 4 ζ(5) = , , , √ 31 − 31 31 558 7 6451 1088 8713 17 ζ(9) = , , , 72571950√7 543 −2172 2172 684521 16480 4219139 515 ζ(13) = , , , 751529653875√7 8239 −1054592 1054592 7556214529 261248 267518969 2041 ζ(17) = , , , 808189287857201250√7 130623 − 66878976 66878976 11042228011 4191904 274720686005 130997 ζ(21) = , , , 115045098786113871375√7 2095951 − 68680122368 68680122368 93518263081637 67120832 35190647292091 ζ(25) = , , , 94909028455546692340078125√7 33560415 − 8797661429760 1048763 8797661429760 Examples of the ﬁrst few ζ(4k+1) generated by Theorem4.5. The coeﬃcients are 4k+1 listed respectively for π , √15π( 4k 1), L √15π( 4k 1), L 2√15π( 4k 1), Se− − − e− − − e− − − L 4√15π( 4k 1). e− − −

5 7 33 1073 33 ζ(5) = , , , , 378√15 √15 16 − 256 256 19 17 513 262913 513 ζ(9) = , , , , 145530√15 7√15 256 − 65536 65536 5623 7 8193 67121153 8193 ζ(13) = , , , , 4214184975√15 33√15 4096 −16777216 16777216 152161 223 131073 17180065793 131073 ζ(17) = , , , , 11136941565750√15 −119√15 65536 − 4294967296 4294967296 2100413011 1673 2097153 4398049656833 ζ(21) = , , , , 15039186678619228125√15 −305√15 1048576 −1099511627776 2097153 1099511627776 368670553 8143 33554433 ζ(25) = , , , √ − √ 16777216 266533834992158608875 15 231 15 1125899957174273 33554433 , − 281474976710656 281474976710656 276635171660523838 30233 536870913 ζ(29) = , , , √ √ 268435456 18471447539635216765490460984375 15 3263 15 288230376957018113 536870913 , − 72057594037927936 72057594037927936 16 SHUBHOBANERJEEANDBLAKEWILKERSON

Examples of the ﬁrst few ζ(4k 1) generated by Theorem5.2. The coeﬃcients are 4k 1 L − L L listed respectively for π − , √7π ( 4k+1), 2√7π ( 4k+1), 4√7π ( 4k+1). e− − e− − e− −

29√7 24 52 6 ζ(3) = , , , " 1980 11 −11 11# 851 240 1927 15 ζ(7) = , , , 963900√7 119 − 476 476 98983 3984 510073 249 ζ(11) = , , , 11006745750√7 1991 −127424 127424 120891949 65712 26916047 4107 ζ(15) = , , , 1310075958262500√7 32855 − 6728704 33643520 304799492533 1050576 34425307261 65661 ζ(19) = , , , 321754984333646613750√7 525287 − 8606302208 8606302208 3069248396337203 16776432 1759136500931 ζ(23) = , , , 315604617827322095616093750√7 8388215 − 439784046592 1048527 2198920232960 Examples of the ﬁrst few ζ(4k 1) generated by Theorem5.3. The coeﬃcients are 4k 1 − L L listed respectively for π − , √15π( 4k+1), √15π( 4k+1), 2√15π( 4k+1), Se− − e− − e− − L 4√15π( 4k + 1). e− −

√15 1 9 77 9 ζ(3) = , , , , " 100 −√15 4 −16 16# 73 11 129 16577 129 ζ(7) = , , , , √ − √ 64 − 4096 4096 56700 15 3 15 82889 61 2049 4197377 2049 ζ(11) = , , , , 6385128750√15 −5√15 1024 −1048576 1048576 3103 11 32769 1073790977 32769 ζ(15) = , , , , √ − √ 16384 − 268435456 268435456 17239847625 15 3 15 269130227 781 524289 274878693377 ζ(19) = , , , , 192947512626618750√15 171√15 262144 − 68719476736 524289 68719476736 247753871371 1451 8388609 70368756760577 ζ(23) = , , , , 17365083188883060881250√15 989√15 4194304 −17592186044416 8388609 17592186044416 Department of Physics, Rhodes College, 2000 N. Parkway, Memphis, TN 38112 E-mail address: [email protected] E-mail address: [email protected]