On Plouffe's Ramanujan Identities

ON PLOUFFE’S RAMANUJAN IDENTITIES LINAS VEPŠTAS A BSTRACT. Recently, Simon Plouffe has discovered a number of identities for the Riemann zeta function at odd integer values. These identities are obtained numerically and are inspired by a prototypical series for Apéry’s constant given by Ramanujan: ζ (3) = ∞ 7π 3 1 − 2 ∑ 3 2π n − 1) 180 n (e n=1 Such sums follow from a general relation given by Ramanujan, which is rediscovered and proved here using complex analytic techniques. The general relation is used to derive many of Plouffe’s identities as corollaries. The resemblance of the general relation to the structure of theta functions and modular forms is briefly sketched. 1. I NTRODUCTION Inspired by an identity for ζ (3) given in Ramanujan’s notebooks [4, chapter 14, formulas 25.1 and 25.3], ∞ 7π 3 1 ζ (3) = − 2 ∑ 3 2πn − 1) 180 n=1 n (e Plouffe describes a set of similar identities[8],[9] that were discovered numerically using arbitrary-precision software. For example, Plouffe gives an identity for ζ (7): ζ (7) = ∞ 19π 7 1 − 2 ∑ 7 2πn 56700 n (e − 1) n=1 This text provides an analytically derived formula for expressions of this type. The resulting general formula, valid for integer m ≥ 1, is ∞ 1 4m−1 (e2πn − 1) n n=1 ζ (4m − 1) = −2 ∑ − 1 (2π)4m−1 2 2m B2 j B4m−2 j ∑ (−1) j (2 j)! (4m − 2 j)! j=0 where Bk is the k’th Bernoulli number. The above is a special case of a yet more general formula, derived and presented in a later section, allowing pairs of such sums to be related. From this, one may obtain expressions such as   37π 3 2 ∞ 1 1 4 + 4πn − ∑ 3 πn ζ (3) = 900 5 n=1 n e − 1 e − 1 There are an (uncountable) infinity of similar sums, each giving a different series summation for ζ (4m − 1). Taking linear combinations of these, one may choose to cancel the zeta Date: 14 May 2006 Revised 27 November 2010. 2000 Mathematics Subject Classification. 11M06. 1 ON PLOUFFE’S RAMANUJAN IDENTITIES 2 terms, to obtain summations for odd powers of π. Thus, for example, combining the above with Ramanujan’s series for Apéry’s constant, one gets:   ∞ 4 1 5 1 π3 = ∑ 3 πn − + 180 n=1 n e − 1 e2πn − 1 e4πn − 1 Again, there are an uncountable infinity of such relations. Plouffe also notes similar relations for the other odd integers; for example, ζ (5) = π5 72 ∞ 1 1 2 ∞ − − ∑ ∑ n5 (e2πn + 1) 294 35 n=1 n5 (e2πn − 1) 35 n=1 The general form for this type of expression may be shown to be   1 + (−4)m − 24m+1 ζ (4m + 1) ∞ 1 4m+1 (e2πn + 1) n n=1 =2∑   ∞ +2 24m+1 − (−4)m ∑ 1 4m+1 (e2πn − 1) n n=1 m +(2π)4m+1 ∑ (−4)m+ j j=0 1 + (2π)4m+1 2 2m+1 B4m−4 j+2 B4 j (4m − 4 j + 2)! (4 j)! B4m−2 j+2 B2 j ∑ (−4) j (4m − 2 j + 2)! (2 j)! j=0 The methods described in this text also allow for a large generalization of these types of sums. Defining ∞ 1 Pk (τ) = ∑ k 2πinτ − 1) n=1 n (e these generalizations follow from a modular equation relating Pk (τ) to Pk (−1/τ) for odd integers k, the derivation and proof of which is the one of the main topics of this note. The modular relation is not new; it appears in Ramanujan’s Notebooks [4, Chapter 14 Entry 21] as ) ( ∞ k2n−1 −n 1 α ζ (2n + 1) + ∑ 2αk 2 −1 k=1 e ( ) ∞ k2n−1 −n 1 = (−β ) ζ (2n + 1) + ∑ 2β k 2 −1 k=1 e − 22n = π2 n+1 B2k B2n+2−2k ∑ (−1)k (2k)! (2n + 2 − 2k)! α n+1−k β k k=0 and n any positive integer. Berndt implies that this forwhere α > 0, β > 0 with αβ mula is the most studied of all the notebooks; it has been independently discovered perhaps a half-dozen times, and proven twice as often. It has been generalized to L-functions, and to rational values of k; Berndt provides a long list[4] of the various proofs and generalizations made. Much of this paper is devoted to (yet another! independently discovered) proof of this relation, followed by a series of lemmas that provide the connection to Plouffe’s results. In searching for curious and interesting special cases of this relation, one senses that only ON PLOUFFE’S RAMANUJAN IDENTITIES 3 the tip of the iceberg has been seen. Unexplored possibilities include, for example, considering τ ∈ Q[i], the field of Gaussian rationals, or from considering Diophantine roots of quadratics. The rest of this paper is roughly laid out as follows: The second section provides a review of previous related results. The third section gives a relationship between the sums and the polylogarithm, and thence to an integral on the complex plane. The fourth section examines the related contour integral, which is easily integrated via Cauchy’s residue theorem to give a finite sum involving the Bernoulli numbers. The fifth section relates the contour integral to the polylogarithm integral, thus resulting in a functional equation for Pk (τ). The sixth section applies the functional equation, providing a variety of lemmas, many of which explain Plouffe’s discoveries. The seventh section give a pair of relationships on the Bernoulli numbers that arise naturally in this context. The eighth section explores the modular nature of the relations on Pk (τ), followed by a conclusion. An appendix gives a derivation of an integral representation of the polylogarithm, that is central to the analysis. 2. R ELATED SUMS A large number of similar sums have been explored before; this section reviews some of these. Perhaps the most forthright is a sum given by Ramanujan in a famous letter to Hardy[7], stating that 213 313 1 113 + 4π + 6π +··· = −1 e −1 e −1 24 A generalization of this sum, e2π ∞ n4k+1 B4k+2 ∑ e2πn − 1 = 4(2k + 1) n=1 is proved by Berndt[2], and attributed to Glaisher[6]. This and many related results are derived by Zucker[11], based on the theory of Jacobian elliptic functions. A similar result is stated by Apostol in the form of an exercise[1, see exercise 15 at end of chapter 1.]: n4k+1 24k+1 − 1 B4k+2 = 1 + enπ 8k + 4 ∞ ∑ odd Many sums resembling those in this note are given by Zucker[12]. Some of these are n=1;n ∞ 1 1 n=1 ∞ ∞ coth(πmx) − 1 m m=1 ∑ n (e2πnx − 1) = 2 ∑ (−1)n 1 ∞ ∑ n (e2πnx − 1) = 2 ∑ (−1)m n=1 m=1 ∞ 1 1 coth(πmx) − 1 m ∞ 1 − tanh(πmx) m m=1 ∑ n (e2πnx + 1) = 2 ∑ n=1 ∞ 1 ∞ 1 − tanh(πmx) (−1)n = ∑ n (e2πnx + 1) 2 ∑ (−1)m m n=1 m=1 The fount of inspiration for such sums is Ramanujan. Sums given in Chapter 14 of Part II of the Ramanujan’s Notebooks[4] include entry 8: ∞ α ∞ β sinh(2αnk) sin(2β nk) α k + β ∑ e2α 2 n − 1 ∑ e2β 2 n − 1 = 4 coth(αk) − 4 cot(β k) − 2 n=1 n=1 ON PLOUFFE’S RAMANUJAN IDENTITIES 4 and another similar one relating cos and cosh. Above, k is any positive integer, αβ = π and 0 < β k < π. Entry 13 generalizes the sums mentioned previously, h iB ∞ ∞ n2k−1 n2k−1 2k = α k − (−β )k − (−β )k ∑ 2β n α k ∑ 2αn − 1 e 4k −1 n=1 n=1 e This time, one takes αβ = π 2 and k > 1 an integer. The above follows from sums on the divisor function, as is frequently noted. Sums involving pairs of Bernoulli numbers also appear in the analysis of the Dedekind eta function. Thus, Sho Iseki’s transformation formula, as described by Apostol[1, see theorem 3.5], is    2  1 2 B2−n (α)Bn (β ) − πz ∑ Λ(α, β , z) = Λ 1 − β , α, n z (iz)n n=0 where Λ is given by ∞ Λ(α, β , z) = ∑ [λ (z(r + α) − iβ ) + λ (z(r + 1 − α) + iβ )] r=0 and ∞ e−2πmx m m=1 A sum linking the Bernoulli and Euler numbers is given by Berndt[3]: √   ∞ n+1 sech (2n − 1)π 3/2 ∑ (−1) (2n − 1)6k+1 n=1 λ (x) = ∑ h 3k 1 πi B6k−2m E2m+1 = (−1)k+1 π 6k+1 ∑ cos (2m + 1) 2 3 m=0 (2m + 1)! (6k − 2m)! Perhaps the results that are closest to those presented in this paper are those noted by Borwein, et al [5, Section 5], and in particular, giving a very similar result involving ζ (4m − 1) and ζ (4m + 1). 3. T HE P OLYLOGARITHM The recurring theme in Plouffe’s identities is the sum ∞ Ss (x) = 1 ∑ ns (exn − 1) n=1 with s usually an odd√positive integer and x = π or x = 2π or possibly other interesting values, such as x = π m for some integer m. This sum may be converted into a sum over polylogarithms, and subsequently into an integral. The integral may, after some difficulties, be converted into a contour integral, whereupon it may be evaluated by Cauchy’s residue theorem. The result is a finite sum whose general structure resembles those of Plouffe’s and Ramanujan’s identities. This section develops the first part of this analysis. To find the polylogarithm, one expands ∞ Ss (x) = ∞ e−xn(m+1) ns n=1 m=0 ∑∑ ∞ = ∑ Lis m=1 e−xm
ON PLOUFFE’S RAMANUJAN IDENTITIES 5 which is generally valid for ℜx > 0. The above is easily obtained by applying the expansion ∞ 1 = ∑ zm 1 − z m=0 and the series definition of the polylogarithm: ∞ Lis (z) = zn ∑ ns n=1 The polylogarithm may be expressed in terms of an integral as Z c+i∞
1 Lis e−u = Γ(z)ζ (z + s)u−z dz 2πi c−i∞ A derivation of this is given in the appendix. Here, Γ(z) = (z − 1)! is the classical Gamma function. The line of integration is taken to be to the right of all of the poles in the integrand, namely, c > 1. Using this in the summation, one obtains Ss (x) = = 1 ∞ ∑ 2πi m=1 1 2πi Z c+i∞ Γ(z)ζ (z + s) (xm)−z dz c−i∞ Z c+i∞ Γ(z) xz c−i∞ ζ (z + s)ζ (z) dz The exchange of the order of summation and integration is justified precisely when one has c > 1. The last integral has poles at z = 1 and z + s = 1 coming from the zeta functions and poles at all of the non-positive integers coming from the Gamma function. The last integral shows that the series is the inverse Mellin transform of Γ(z)ζ (z + s)ζ (z). If the integral can somehow be converted into a closed contour on the left, then it may be evaluated in a straight-forward way by means of Cauchy’s residue theorem. Performing this closure is in fact harder than one might hope, as there are non-zero contributions to the contour from its closure. The next section evaluates the Cauchy integral, assuming that the contour can be closed. The section after that computes the contributions from closing the contour integral. Upon doing this, Plouffe’s identities, and many more, become available. 4. T HE C ONTOUR I NTEGRAL Define the contour integral as Is (x) = 1 2πi I γ Γ(z) ζ (z + s)ζ (z)dz xz where the contour γ encircles the poles at z = 1, z + s = 1 and z = 0, −1, −2, . . . in the usual, right-handed fashion. Then, one uses Cauchy’s theorem, which states that f (a) = 1 2πi for simple poles, and that 1 f (a) = 2πi ′ I I f (z) dz z−a f (z) dz (z − a)2 for double poles. For the pole at z = 1, one obtains the residue Res(z = 1) = ζ (s + 1) x ON PLOUFFE’S RAMANUJAN IDENTITIES 6 For the poles at z = −n, one obtains the residue Res(z = −n) = (−x)n ζ (s − n)ζ (−n) n! and so one has Is (x) = Res(z = 1 − s) + ∞ ζ (s + 1) (−x)n +∑ ζ (s − n)ζ (−n) x n=0 n! For s not an integer, one has Γ(1 − s) ζ (1 − s) x1−s However, the interesting case is for s = k a positive integer. In this case, the pole overlays another pole from the Gamma, and one has a double pole. This is just a little trickier to evaluate: I I 1 Γ(z) f (z) 1 ζ (z + s)ζ (z)dz = dz z 2πi z=1−k x 2πi z=1−k (z + k − 1)2   ζ (z) d 2 (z + k − 1) Γ(z) ζ (z + k) z = dz x z=1−k Res(z = 1 − s) = To perform the derivative, one will need to use the identities d (s − 1)ζ (s) ds =γ s=1 where γ = 0.577 . . . is the Euler-Mascheroni constant, and d (z + n)Γ(z) dz = (−1)n z=−n ψ(n + 1) Hn − γ = (−1)n Γ(n + 1) n! where ψ(z) is the digamma function, and Hn is the n’th harmonic number. Putting these together, one obtains 1 2πi  (−x)k−1  ′ Γ(z) ζ (z + s)ζ (z)dz = ζ (1 − k) + (Hk−1 − ln 2π) ζ (1 − k) z (k − 1)! z=1−k x I Adding this to the other contributions, one gets Ik (x) = ζ (k + 1) + x + ∞ ∑ n=0 n 6= k − 1 (−x)n ζ (k − n)ζ (−n) n!  (−x)k−1  ′ ζ (1 − k) + (Hk−1 − ln 2π) ζ (1 − k) (k − 1)! When k is an odd integer, the above simplifies in two ways. First, ζ (1 − k) vanishes, because the zeta function vanishes at all negative even integers. Similarly, the infinite sum becomes finite: when k is an odd integer, one has either ζ (k − n) = 0 or ζ (−n) = 0 for all n > k. Thus, for k an odd integer, one has Ik (x) = ζ (k + 1) + x k ∑ n=0 n 6= k − 1 (−x)k−1 ′ (−x)n ζ (k − n)ζ (−n) + ζ (1 − k) n! (k − 1)! ON PLOUFFE’S RAMANUJAN IDENTITIES 7 For the remainder of the paper, it is assumed that, in this context, k > 1 is an odd integer, unless explicitly stated otherwise. The above evaluation of the contour integral is the main result of this section. To see that this is a key result, one may substitute k = 3 and x = 2π to obtain 7π 3 I3 (2π) = −ζ (3) + 180 which should be recognizable as a portion of Ramanujan’s identity. For k = 7, one has 19π 7 56700 which resembles one of the results given by Plouffe. To complete the connection, one must relate the contour integral Ik (x) to the sum Sk (x). This is done in the next section. First, however, to drive the point home, one must observe that most of the terms in the above expression are rational multiples of powers of π. This follows from the zeta function being related to the Bernoulli numbers Bn at even integers: I7 (2π) = −ζ (7) + ζ (2n) = (−1)n+1 (2π)2n B2n 2(2n)! for integer n ≥ 0. At the negative values, one has Bn+1 ζ (−n) = − n+1 while the derivative is (2n)! ζ ′ (−2n) = (−1)n ζ (2n + 1) 2(2π)2n for integer n > 1. Using these in the above expression for Ik (x), and rearranging terms a bit, one gets   x k−1  −2Ik (x) = ζ (k) 1 − 2πi  (k+1)/2  1 k+1 x2 j (2πi)k+1−2 j B2 j Bk+1−2 j + ∑ 2j x (k + 1)! j=0 The above introduces the binomial coefficient   n! n = k k!(n − k)! √ Although the imaginary number i = −1 appears in the above, it is always squared, and thus is just a sign-keeping device. Every term in the sum is purely real. When the contour integral is written in this form, it may now be seen that for x being any rational multiple of π, that is, x = pπ/q for any integers p, q, that the coefficient of ζ (k) is a rational number, and that the second term is another rational times π k . p A further curiosity in this regard is noted by Plouffe: if one takes x = π p/q for integers p and q, one also gets simple expressions: because k is odd, the coefficient of ζ (k) p k For rational x, one is still a rational, and the coefficient of π is p/q times some rational. p still has that the sum is a rational polynomial in π 2 , and for x = π p/q, one still has that the sum is a rational polynomial in π. The ocean-full of rationals here suggest that some sort of p-adic analysis might be interesting. The appearance of the square root suggests that there is a relation to complex multiplication, or that one may have interesting results on the field of Gaussian integers. ON PLOUFFE’S RAMANUJAN IDENTITIES 8 5. E VALUATING THE C ONTOUR I NTEGRAL The goal of this section is to relate the sum Sk (x) to the contour integral Ik (x). Theorem 1. For odd integers k, one has that (k−1)/2 Sk (x) = Ik (x) + (−1)  x k−1  4π 2  Sk 2π x For the remainder of this text, this will be referred to as the “functional equation for Sk ”. Proof. To prove this result, consider evaluating the contour integral Is (x) for a tall rectangular contour surrounding the poles at z = 1, 0, −1, . . . , 1 − s. Thus, write Is (x) = A + B + C + D with A being the integral from c − ih to c + ih for a constant c > 1 and the height h large, eventually taking the limit h → ∞. That is, A forms the right hand side of the rectangular contour. In the limit of h → ∞, one has by definition A = Ss (x) Let B and C be the top and bottom of the contour, so that for B, the integral runs from c + ih to ih − s − ε leftwards. For C, the integral runs rightwards from −ih − s − ε to c − ih; here we take ε > 0. The integral D on the left hand side of the rectangle closes the contour, running downwards, from ih − s − ε to −ih − s − ε. The integrals B and C will vanish in the limit of the height h → ∞. This can be easily seen after a simple change of variable: ih−s−ε Γ(z) 1 ζ (s + z)ζ (z) dz 2πi ih+c xz Z −s−ε 1 Γ(u + ih) = ζ (s + u + ih)ζ (u + ih) du 2πi c xu+ih Much of the integrand is O(1) in h, or polynomially thereabouts. The integrand is dominated by the Gamma function, which, from Stirling’s approximation, may be seen to be   Γ(u + ih) = O e−πh/2 Z B = The complex conjugate argument applies to C, and thus B and C vanish in the limit of h → ∞. To evaluate D, begin by writing ih−s−ε Γ(z) 1 ζ (s + z)ζ (z) dz 2πi −ih−s−ε xz Z 1 s+ε ih u x Γ(−u − s − ε)ζ (−u − ε)ζ (−u − s − ε) du x = − 2πi −ih after a change of variable z = −u − s − ε. One then applies the functional equations for Gamma: π Γ(1 − z)Γ(z) = sin πz and for zeta: πs ζ (s) = 2s π s−1 sin Γ(1 − s)ζ (1 − s) 2 to obtain  u sin u+ε π sin u+s+ε π

x x s+ε 1 ε R ih 2 2 D = 2π12 i 2π × −ih 2π 4π 2 sin(u + s + ε)π Γ(1 + u + ε)ζ (1 + u + ε)ζ (1 + s + u + ε) du D = − Z ON PLOUFFE’S RAMANUJAN IDENTITIES 9 Another change of variable, this time as w = 1 + u + ε, takes the integral to a slightly more recognizable form: Z 1  x s−1 1+ε+ih  x w cos w2 π cos w+s 2 π D=− Γ(w)ζ (w)ζ (w + s) dw 2 πi 2π 4π sin(w + s)π 1+ε−ih Next, by taking s = k to be an odd integer, the trigonometric piece simplifies and looses its w dependence: cos w2 π cos w+k 1 2 π = (−1)(k+1)/2 sin(w + k)π 2 Pulling out this piece, one regains a recognizable integral, so that, in the limit h → ∞, one finally reaches the claimed result:  x k−1  4π 2  Sk D = (−1)(k+1)/2 2π x that is, (k−1)/2 Ik (x) = A + D = Sk (x) − (−1) for odd integer k.  x k−1  4π 2  Sk 2π x  The next section will review the application of this and the preceding sections to specific, simple values of x, thus regaining many of Plouffe’s sums. 6. L EMMAS AND A PPLICATIONS The first corollary demonstrates Plouffe’s simplest sums for ζ (4m − 1). Corollary 2. For m integer, one has I4m−1 (2π) = 2S4m−1 (2π). Proof. Substitute x = 2π in the functional equation.  This corollary provides the first concrete result of this exposition, namely that   ∞ 1 (2π)4m−1 2m 4m 2 ∑ 4m−1 2πn = −ζ (4m − 1) + ∑ 2 j (−1) j B2 j B4m−2 j n (e (4m)! − 1) j=0 n=1 which completely resolves one set of relationships given by Plouffe. The functional equation opens additional possibilities. By substituting x = 2π p/q, one obtains, for k = 4m − 1, that   ∞ ∞ 1 2π p 1 k−1 k−1 k−1

q ∑ k 2π pn/q − 1 + p ∑ nk e2πqn/p − 1 = q Ik q n=1 n e n=1 Thus, for example, by choosing p = 2, q = 1 and k = 3, on obtains   1 37π 3 2 ∞ 1 4 + 4πn ζ (3) = − ∑ 3 πn 900 5 n=1 n e − 1 e − 1 and similarly, for k = 7,   1 409π 7 2 ∞ 1 4 + ζ (7) = − ∑ 7 πn 94500 5 n=1 n e − 1 e4πn − 1 and one may proceed in a similar manner. There are an uncountable infinity of such relations (since p/q need not be rational). One may take arbitrary linear combinations of ON PLOUFFE’S RAMANUJAN IDENTITIES 10 these; or if one desires, one may subtract to cancel out zeta terms, leaving behind an (uncountable) infinity of relations for powers of π. The next corollary shows that something more is needed for ζ (5), ζ (9), and so on, since the most direct approach does not give any information for such sums. Corollary 3. For k an odd integer, one has (k+1)/2 Ik (x) = (−1)  4π 2 x k−1  2  4π Ik x Proof. This may be proved by applying the functional equation twice in a row. That is, it may be proved by substituting x → 4π 2 /x in the functional equation and then employing the result.  Corollary 4. For m a positive integer, one has I4m+1 (2π) = 0. Proof. Substitute x = 2π in the preceding corollary.  Following directly from the above is another corollary: Corollary 5. For m a positive integer, one has 2m+1 0= B4m−2 j+2 B2 j ∑ (−1) j (4m − 2 j + 2)! (2 j)! j=0 Proof. Write out the value of I4m+1 (2π) in detail.  From the above, it should be clear that the functional equation for Sk does not provide any statements about Sk when k = 4m + 1. To obtain results on sums involving k = 4m + 1, one must introduce ∞ 1 Ts (x) = ∑ s xn n=1 n (e + 1) Theorem 6. One has Ts (x) = Ss (x) − 2Ss (2x) Proof. This may be proved by re-writing in terms of the polylogarithm, along the lines of the earlier development: ∞
Ts (x) = − ∑ (−1)m Lis e−xm m=1 The even and odd terms are regrouped, as ∞ Ts (x) = ∑ Lis m=1 which is seen to be a sum of Ss ’s.

e−xm − 2Lis e−2xm The results for ζ (5), etc. follow from a critical observation: that Ss (x + 2πi) = Ss (x) is a periodic function. This periodicity is employed directly in the next theorem.  ON PLOUFFE’S RAMANUJAN IDENTITIES 11 Theorem 7. For positive integer m, one has S4m+1 (2π) = I4m+1 (2π(1 + i))   T4m+1 (2π) m m + 2 · 4 S (2π) − 4 I (π) +(−1)m 4m+1 4m+1 4m Proof. Using periodicity, one writes, for k = 4m + 1, Sk (2π + 2πi) = Sk (2π) = Ik (2π(1 + i)) + 2(k−1)/2 e3iπ(k−1)/4 Sk (π(1 − i)) The series Sk (π(1 − i)) doesn’t have an imaginary part; rather, it is an alternating series, which may be expanded and written as Sk (π(1 − i)) = −Tk (π) + Tk (2π) + Sk (2π) 2k The Tk (π) term may be eliminated by writing Tk (π) = Sk (π) − 2Sk (2π) and the Sk (π) term may be eliminated by Sk (π) = Ik (π) + 21−k Sk (4π) = Ik (π) + 2−k [Sk (2π) − Tk (2π)] Performing the various substitutions suggested above proves the theorem.  As an example of the application of the above theorem, take m = 1, that is, k = 5. One easily finds that I5 (π) = − and that Combining these, one gets π5 15 ζ (5) + 32 9 · 64 π5 5 I5 (2π(1 + i)) = − ζ (5) + 2 9 · 15 ζ (5) = π5 2 − [T5 (2π) + 36S5 (2π)] 294 35 which is given by Plouffe. The theorem may be used to generate similar expressions for all ζ (4m + 1). Curiously, the theorem yields results for m = 0 as well. In this case, one finds S1 (2π) + T1 (2π) = π 3 − log 2 6 4 Many identities for π are possible by taking two different expressions for a given zeta, and subtracting them, leaving behind a rational combination of the sums and π. Thus, for example, the following theorem for Apéry’s constant: Theorem 8. A series expression for π 3 is given by π 3 = 720 · S3 (π) − 900 · S3 (2π) + 180 · S3 (4π) ON PLOUFFE’S RAMANUJAN IDENTITIES 12 Proof. This follows by taking the general expression for k = 4m − 1: 16m S4m−1 (π) + 4S4m−1 (4π) = 16m I4m−1 (π) 1 = − ζ (4m − 1) [16m + 4] 2 2m B4m−2 j B2 j −(2π)4m−1 ∑ (−4) j (4m − 2 j)! (2 j)! j=0 solving for ζ (4m − 1) and then using 2m 2S4m−1 (2π) = I4m−1 (2π) = −ζ (4m − 1) + (2π)4m−1 B4m−2 j B2 j ∑ (−1) j (4m − 2 j)! (2 j)! j=0 to eliminate the appearance of the zeta. The resulting expression may then be solved for π 4m−1 .  Relationships involving square roots also arise naturally. Theorem 9. One has √ ! 2π 3 3 5π 3 1  √  3 − S3 2π 3 − S3 ζ (3) = 72 2 2 Proof. This follows from the general expression given earlier, that    k−1     2π p p 2πq 2π p Sk + Sk = Ik q q p q √ Here, making the substitution q = p one obtains  √  h i 2π p 1 √ (k−1)/2 Sk Sk (2π p) + p = − ζ (k) 1 − (−p)(k−1)/2 p 2 + (−1)(k−1)/2 (2π)k √ 2 p (k+1)/2 ∑ (−p) j j=0 Bk+1−2 j B2 j (k + 1 − 2 j)! (2 j)! The specific result follows after choosing k = 3 and p = 3.  7. S OME B ERNOULLI NUMBER IDENTITIES In addition to the previously noted identity 2m+1 0= B4m−2 j+2 B2 j ∑ (−1) j (4m − 2 j + 2)! (2 j)! j=0 there are several other identities on sums of Bernoulli numbers that result from the previous developments. These are briefly stated here. Theorem 10. For integer m, one has   m B4m−4 j B4 j B4 j+2 B4m−4 j+2 +2 0 = ∑ (−4) j (4m − 4 j + 2)! (4 j)! (4m − 4 j)! (4 j + 2)! j=0 Proof. Consider the sums resulting from 0 = I4m+1 (2π(1 + i)) − I4m+1 (2π(1 − i)).  ON PLOUFFE’S RAMANUJAN IDENTITIES 13 Theorem 11. For integer m, one has 2m B4m−2k B2k ∑ (−1)k (4m − 2k)! (2k)! m = B4m−4 j B4 j ∑ (−4) j (4m − 4 j)! (4 j)! j=0 k=0 m−1 −2 B4m−4 j−2 B4 j+2 ∑ (−4) j (4m − 4 j − 2)! (4 j + 2)! j=0 Proof. Consider the sums resulting from the identity I4m−1 (2π) = I4m−1 (2π(1 + i)) + I4m−1 (2π(1 − i))  8. M ODULAR R ELATIONS The various sums and quantities above can be seen to be quasi-modular by making a simple change of variable, namely by making the substitution x = 2πiτ. By “quasimodular”, it is meant that the various terms almost have simple behaviors under the Mobius transformation τ → (aτ +b)/(cτ +d) for integer a, b, c and d. In this respect, the sums bear close resemblance to theta functions, which obey similar relations. These relationships are brought to focus here. First, define Kk (τ) = Ik (2πiτ). This change of variable results in a definition which seems simpler than that for Ik . An expansion in τ is often referred to as a “Fourier series” in the context of hyperbolic geometry: Kk (τ) = τ k−1 − 1 (2πi)k ζ (k) − 2 2τ (k+1)/2 ∑ τ2 j j=0 Bk+1−2 j B2 j (2 j)! (k + 1 − 2 j)! This quantity is almost a modular form of weight k − 1, in that   −1 Kk = −τ 1−k Kk (τ) τ Its only “almost” a modular form, because it is not periodic in τ, that is Kk (τ + 1) 6= Kk (τ) By contrast, Pk (τ) = Sk (2πiτ) is periodic: Pk (τ + 1) = Pk (τ) but is not quite modular under inversion:     −1 −1 Pk = τ 1−k Pk (τ) + Kk τ τ One may define a simple variant of the sums that does have a simple transformation under inversion, namely 1 Mk (τ) = Pk (τ) − Kk (τ) 2 which transforms as   −1 = −τ 1−k Mk (τ) Mk τ However, Mk is then not periodic. ON PLOUFFE’S RAMANUJAN IDENTITIES 14 F IGURE 8.1. Phase of P−1 (q) on the unit disk This graphic shows the phase ∞ n n −1 q n=1 arg P−1 (q) = arg ∑ where arg f (z) = ℑ log f (z) is the usual arg of a function. The color scheme is such that black represents areas where arg > 0 and red represents areas where arg <0. The absence of other colors indicates that the phase is rather closely confined to the vicinity of 0 for most all of the disk. Numerically, the absolute value of the phase is smaller than 10−3 for much of the disk. In particular, this indicates that there are no zeros at all in the interior of the disk, as a zero would be surrounded by a region where the phase wraps around by 2π. The function Ps (q) does have poles at q = e2πim/n for all rationals m/n; these are visible at the edges of the disk. The fractal nature of this image is the characteristic signature of a modular form of weight 2; the self-similar regions are just copies of the fundamental region of the modular group SL(2, Z). ON PLOUFFE’S RAMANUJAN IDENTITIES 15 F IGURE 8.2. Phase of P−5 (q) on the unit disk This graphic shows the phase n5 n n=1 q − 1 ∞ arg P−5 (q) = arg ∑ The color scheme is as in the previous image, and similar conclusions apply: there are no zeros at all in the interior of the disk. The absolute value of the phase is tiny: numerically, it is within 10−9 of zero for much of the disk. The fractal nature of this image is the characteristic signature of a modular form of weight 6; it should be compared to the image of the modular invariant g3 shown in the next image. The analytic structure of Pk is curious: it has a pole at τ = i∞ and. by the inversion formula and periodicity, at every rational value of τ. This is clearly visible in the graphic ON PLOUFFE’S RAMANUJAN IDENTITIES 16 F IGURE 8.3. Graph of the modular invariant g3 . This figure shows the imaginary part of the modular invariant g3 , one of the invariants of an elliptic curve. Specifically, it shows the imaginary part of # " ∞ 8π 6 n5 qn g3 (q) = 1 − 504 ∑ n 27 n=1 1 − q which is a modular form of weight 6. The colors are chosen such that black represents areas that are negative, blue and green represent areas with smaller values, and red those areas with the largest values. ON PLOUFFE’S RAMANUJAN IDENTITIES 17 below, which shows P3 on the q-series or “punctured disk” coordinates q = e2πiτ : ∞ Ps (q) = 1 ∑ ns (qn − 1) n=1 In this form, the relation to modular forms becomes the most immediate. Consider the Lambert series: ∞ ∞ 1 qn ∑ ns 1 − qn = −ζ (s) − Ps (q) = ∑ σ−s (m)qm n=1 m=1 Here, σs (m) is the divisor function: σs (m) = ∑ ns n|m with the notation n|m denoting that the sum extends over all divisors n of m. This should be compared to the Eisenstein series [1, see section 3.10] G2k (q) = 2ζ (2k) + 2(2πi)2k ∞ ∑ σ2k−1 (m)qm (2k − 1)! m=1 That is, the sums Ps can be re-written in terms of the Eisenstein series G1−s , and the behaviour of one under the action of the modular group can be igven in terms of the other. 9. C ONCLUSIONS Most of the sums discussed here are suggestive of linear algebra. So for example, one may write the sum Sk as the dot-product between the (infinite-dimensional) vector n−k
−1 and the vector e2πn − 1 . The evaluation of these sums forms a regular pattern in k, suggesting that, for example, that n−k could be taken to be the matrix elements of some linear operator. However, the significance of this operator (aside from assorted shallow results and relations) is completely unclear. The sums over the Bernoulli numbers are even reminiscent of some crazy Atiyah-Singer-like indexing; but the underlying operators are utterly unclear. Put another way, it is well-known in physics and mathematics that regular patterns are the result of symmetries; the sums discussed here form a regular pattern, but the nature of the symmetry that generates it is unclear. During manipulations, one gets the sense that there are plenty of other relations to be discovered; certainly, the mere heft of Ramanujan’s tomes suggest as much. Yet, there is no picture of a generator that can be operated to generate the myriad of relations; some symmetry group presentation seems to be missing. A similar problem exists in the theory of hypergeometric series, where there is an embarrassment of riches in terms of relations and identities, and yet a unifying theory is lacking. The ingredients to the sums discussed here include the Gamma function and the Bernoulli polynomials; among many other properties, these have a common set of relations in the p-adic “multiplication theorems”; the sums here vaguely resemble the multiplication theorems of characteristic zero. There are also similar phenomena and sums that occur in the theory of dynamical systems, and in particular, in symbolic dynamics; there, a group structure, or at least, a monoid structure, together with an explicit treatment in terms of linear operators, is more common. Any of these connections present intriguing avenues for future research; however, the overall problem, of discovering the underlying symmetry that leads to such relations, seems unattainably hard to solve. Thanks to Simon Plouffe for generating interest in such sums. ON PLOUFFE’S RAMANUJAN IDENTITIES 18 10. A PPENDIX : P OLYLOG I NTEGRAL This appendix proves the following theorem: Theorem 12. The polylogarithm may be written as the integral Z c+i∞
1 Γ(r)ζ (r + s)u−r dr Lis e−u = 2πi c−i∞ Proof. The proof below is cribbed from the Wikipedia article on polylogarithms[10]. One begins by writing the the Mellin transform of the polylog, as Z ∞
du Ms (r) = Lis ye−u ur u 0 Using an integral representation of the polylogarithm, Lis (w) = 1 Γ(s) and substituting, one obtains Ms (r) = 1 Γ(s) Z ∞ 0 Z ∞Z ∞ 0 0 t s−1 dt w−1 et − 1 ur−1t s−1 dt du y−1 et+u − 1 A change of variable t = ab and u = a(1 − b) with dt du = a da db gives 1 1 bs−1 (1 − b)r−1 db Γ(s) 0 = Γ(r) Lir+s (y) Ms (r) = Z Z ∞ ar+s−1 0 y−1 ea − 1 da The inverse Mellin transform may now be employed to write Z c+i∞
1 u−r Γ(r) Lir+s (y) dr Lis ye−u = 2πi c−i∞ By setting y = 1, one then uses Lis+r (1) = ζ (s + r) to obtain the desired result.  R EFERENCES [1] Tom M. Apostol. Modular Functions and Dirichlet Series in Number Theory. Springer, 2nd ed. edition, 1990. [2] Bruce C. Berndt. Modular transformations and generalizations of several formulae of ramanujan. Rocky Mountain Journal of Mathematics, 7:147–189, 1977. [3] Bruce C. Berndt. Analytic eisenstein series, theta functions and series relations in the spirit of ramanujan. J. Riene Angew. Math., 303/304:332–365, 1978. [4] Bruce C. Berndt. Ramanujan’s Notebooks, Part II. Springer-Verlag, 1989. ISBN 0-387-96794-X. [5] Jonathan M. Borwein, David M. Bradley, and Richard E. Crandall. Computational strategies for the riemann zeta function. Journal of Computational and Applied Mathematics, 121:247–296, 2000. http://www.maths.ex.ac.uk/ mwatkins/zeta/borwein1.pdf. [6] J. W. L. Glaisher. On the series which represent the twelve elliptic and the four zeta functions. Mess. Math., 18:1–84, 1889. [7] G. H. Hardy. Obituary notice. Proceedings of the Royal Society, Series A, 99:xiii–xxix, 1921. [8] Simon Plouffe. Identities inspired from ramanujan notebooks ii. http://www.lacim.uqam.ca/ plouffe/identities.html, July 21 1998. [9] Simon Plouffe. Identities inspired by ramanujan notebooks (part 2). http://www.lacim.uqam.ca/ plouffe/inspired2.pdf, April 2006. [10] Wikipedia. Polylogarithm, 2006. http://en.wikipedia.org/wiki/Polylogarithm. [11] I. J. Zucker. The summation of series of hyperbolic functions. SIAM Journal of Mathematical Analysis, 10(1):192–206, January 1979. [12] I. J. Zucker. Some infinite series of exponential and hyperbolic functions. SIAM Journal of Mathematical Analysis, 15(2):406–413, March 1984. ON PLOUFFE’S RAMANUJAN IDENTITIES – – E-mail address: ❁❧✐♥❛s✈❡♣st❛s❅❣♠❛✐❧✳❝♦♠❃ 19