Sums of Generalized Harmonic Series (NOT Multiple Zeta Values)


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1 Sums of Series (NOT Multiple Zeta Values) Michael E. Hoffman USNA Basic Notions 6 February 2014
2 1 2 3
3 harmonic series The generalized harmonic series we will be concerned with are H(a 1, a 2,..., a k ) = n=1 1 n a 1 (n + 1) a 2 (n + k 1) a k, where a 1,..., a k are nonnegative integers whose sum is at least 2. We call this quantity an Hseries of length k and weight a a k. Of course Hseries of length 1 are just values of the Riemann zeta function. We will prove that H(a 1,..., a k ) = kζ(m), (1) a 1 + +a k =m where m 2 and the sum is over ktuples of nonnegative integers. This was conjectured by C. Moen last spring, and proved by both of us over the summer.
4 First cases Equation (1) is only interesting if k > 1. The first case is H(2, 0) + H(1, 1) + H(0, 2) = 2ζ(2), which is pretty trivial: evidently H(2, 0) = ζ(2) and H(0, 2) = ζ(2) 1, and the usual telescopingsum argument shows 1 H(1, 1) = n(n + 1) = 1. n=1 More generally, equation (1) for k = 2 is H(m, 0) + H(m 1, 1) + + H(1, m 1) + H(0, m) = 2ζ(m), which reduces to H(m 1, 1) + + H(1, m 1) = 1. (2)
5 Multiple zeta values Equation (1) is superficially similar to the sum theorem for multiple zeta values (conjectured by C. Moen 1988, proved by A. Granville 1997): for m 2, where a 1 + +a k =m, a i 1, a 1 2 ζ(a 1, a 2..., a k ) = ζ(a 1, a 2,..., a k ) = ζ(m) (3) n 1 >n 2 > >n k 1 1 n a 1 1 na 2 2 na k k. But note that the multiple zeta value ζ(a 1,..., a k ) is a kfold sum, while H(a 1,..., a k ) is a simple sum. Also, the lefthand side of (3) is a sum over ktuples of positive integers, while that of (1) is a sum over ktuples of nonnegative integers.
6 Another sum of Hseries We could ask what happens if we sum Hseries over all ktuples of positive integers with sum m 2. There is a nice formula in this case, though by no means as simple as equation (1): a 1 + +a k =m, a i 1 H(a 1,..., a k ) = 1 (k 1)! k 2 ( ) k 2 i i=0 ( 1) i. (4) (i + 1) m k+1 Note that the righthand side is a rational number. The case k = 2 is equation (2).
7 Telescoping We shall build up a chain of lemmas on Hseries. Using 1 n(n+1) = 1 n 1 n+1, note that for a, b 1, H(a, b) = [ n=1 Hence n=1 1 n a (n + 1) b = 1 n a (n + 1) b 1 1 n a 1 (n + 1) b ] = H(a, b 1) H(a 1, b). H(m 1, 1) + H(m 2, 2) + + H(1, m 1) = H(m 1, 0) H(m 2, 1) + H(m 2, 1) H(m 3, 2) + +H(1, m 2) H(0, m 1) = H(m 1, 0) H(0, m 1) = 1, proving equation (2).
8 Sum over indices of fixed sum, fixed positions This generalizes as follows. Lemma (1) Let 1 i < j p, and let a 1,..., a i 1, a i+1,..., a j 1, a j+1,..., a p be fixed nonnegative integers. Then m 1 k=1 H(a 1,..., a i 1, k, a i+1,..., a j 1, m k, a j+1,..., a p ) = 1 j i [H(a 1,..., a i 1, m 1, a i+1,..., a j 1, 0, a j+1,..., a p ) H(a 1,..., a i 1, 0, a i+1,..., a j 1, m 1, a j+1,..., a p )].
9 Sum over indices with fixed nonzero positions We can use Lemma (1) to obtain a formula for the sum of all Hseries of fixed length and weight and nonzero entries at specified positions. Let H p,q (r) = q j=p j r for p q. Lemma (2) For k 1 and 1 i 0 < i 1 < < i k n, let P(i 0,..., i k ) be the set of strings (a 1,..., a n ) of nonnegative integers with a a n = m and a j 0 iff j = i q for some q. Then (a 1,...,a n) P(i 0,...,i k ) k j=1 H(a 1,..., a n ) = ( 1) j 1 H (m k) i 0,i j 1 (i j i 0 )(i j i 1 ) (i j i j 1 )(i j+1 i j ) (i k i j ).
10 Sum over indices with fixed number of nonzero positions For 0 k n 1, let C(k, n; m) be the sum of all Hseries H(a 1..., a n ) of length n, weight m, and exactly k + 1 of the a i nonzero. If k 1, C(k, n; m) is the sum over all sequences 1 i 0 < < i k n of the sums given by the preceding result; since each H (m k) i 0,i j 1 in that result is a sum of 1 with p m k 1 i 0 p i j 1 n 1, we have C(k, n; m) = n 1 j=1 c (n) k,j j m k (5) with c (n) k,j rational. We can deduce results about the c (n) k,j Lemma (2). from
11 An (anti)symmetry property Lemma (3) For 1 k, j n 1, c (n) k,n j = ( 1)k 1 c (n) k,j. Proof. For 1 i 0 < < i k n, let S m n (i 0,..., i k ) be the lefthand side of the equation in Lemma (2). Then if S m n (i 0,..., i k ) = p 1 + p 2 2 m k + + p n 1 (n 1) m k, it follows from Lemma (2) that ( 1) k 1 S m n (n + 1 i k,..., n + 1 i 0 ) = p n 1 + p n 2 2 m k + + p 1 (n 1) m k.
12 Some explicit coefficients Here are some of the coefficients c (n) k,j. c (3) 1,1 = 3 2 c (3) 1,2 = 3 2 c (3) 2,1 = 1 2 c (3) 2,2 = 1 2 c (4) 1,1 = 11 6 c (4) 1,2 = 7 3 c (4) 1,3 = 11 6 c (4) 2,1 = 1 c(4) 2,2 = 0 c(4) 2,3 = 1 c (4) 3,1 = 1 6 c (4) 3,2 = 2 6 c (4) 3,3 = 1 6 1,1 = ,2 = ,3 = ,4 = ,1 = ,2 = 5 8 2,3 = 5 8 2,4 = ,1 = ,2 = ,3 = ,4 = ,1 = ,2 = ,3 = ,4 = 1 24
13 More explicit coefficients For brevity, we give the matrices [(n 1)!c (n) i,j ] for n = 6, Note that all row sums except the first are 0.
14 The first columns the of the matrices above turn out to be wellknown. Let [ n k] be the number of permutations of {1, 2,..., n} with exactly k disjoint cycles; this is the unsigned number of. We set [ 0 0] = 1, and for n 1 the number [ n k] is only nonzero if 1 k n. Some useful facts are [ ] [ ] [ ] ( ) n n n n = (n 1)!, = 1, =. 1 n n 1 2 One has the generating function x(x + 1) (x + n 1) = n k=1 [ ] n x k, n 1. (6) k
15 cont d The following result relates the numbers to the c (n) k,1. Lemma (4) For 1 k n 1, Proof. c (n) k,1 = ( 1)k 1 c k,n 1 = [ ] 1 n. (n 1)! k + 1 By Lemma (3), it suffices to prove the last equality. Lemma (2) says c (n) k,n 1 = ( 1) k 1 (n i 0 ) (n i k 1 ) 1 i 0 < <i k 1 <n
16 cont d Proof cont d. so it s enough to show 1 i 0 < <i k 1 <n (n 1)! (n i 0 ) (n i k 1 ) = [ ] n. k + 1 Now the lefthand side is evidently the sum of all products of n k 1 distinct factors from the set {1, 2..., n 1}, and that this is [ n k+1] follows from consideration of the generating function (6).
17 A general formula The preceding result generalizes as follows. Lemma (5) For 1 k, j n 1, c (n) k,j = q q=1 p=1 ( 1) p 1[ ][ q n+1 q p k+2 p (q 1)!(n q)! We only sketch the proof. The lemma amounts to c (n) k,j c (n) k,j 1 = ( 1) p 1[ ][ j n+1 j ] p k+2 p. (7) (q 1)!(n q)! p=1 ]. By Lemma (3), we can make this about c (n) k,n j c(n) k,n j+1.
18 A general formula cont d Now think of the c (n) k,i as sums of terms via Lemma (2). If j k, then all terms contributing to c (n) k,n j+1, and in addition there are j classes of terms that c (n) k,n j also contribute to contribute to c (n) k,n j and not to c (n) k,n j+1 : these j classes of terms can be matched up with the j terms on the righthand side of equation (7). If j > k, then there are also terms that contribute to c (n) k,n j+1 and not to c(n) k,n j : one can use Lemmas (2) and (4) to match up their sum with a term on the righthand side of equation (7).
19 second sum formula Now we ll prove the (1) and (4). As a warmup, we ll do the second of these first. Replacing k by n, equation (4) is a 1 + +a n=m, a i 1 H(a 1,..., a n ) = 1 (n 1)! n 1 j=1 ( ) n 2 ( 1) j 1 j 1 j m n+1. The lefthand side is C(n 1, n; m), so it suffices to show that ( ) c (n) n 1,j = ( 1)j 1 n 2. (n 1)! j 1
20 second sum formula cont d Since [ n k] = 0 when k > n and [ n n] = 1, this turns out to be an easy consequence of Lemma (5): c (n) n 1,j = = q q=1 p=1 q=1 = 1 (n 1)! = ( 1)j 1 (n 1)! ( 1) p 1[ ][ q n+1 q p n+1 p (q 1)!(n q)! ] ( 1) q 1[ q q ][ n+1 q n+1 q (q 1)!(n q)! ( ) n 1 ( 1) q 1 q 1 q=1 ( ) n 2. j 1 ]
21 first sum formula Next we attack equation (1). We have a 1 + +a n=m Now C(0, n; m) is n 1 H(a 1,..., a n ) = C(0, n; m) + C(k, n; m). k=1 H(m, 0,..., 0) + H(0, m, 0,..., 0) + + H(0,..., 0, m) = ζ(m) + ζ(m) ζ(m) m 1 (n 1) m = nζ(m) (n 1) n 2 2 m 1 (n 1) m n 1 = nζ(m) j=1 n j j m
22 first sum formula cont d so (using equation (5)) a 1 + +a n=m n 1 H(a 1,..., a n ) = nζ(m) Hence, to prove the result requires n 1 j k c (n) k,j k=1 j=1 = n j n j j m n 1 n 1 + k=1 j=1 c (n) k,j j m k. for 1 j n 1. In view of Lemma (5), this is n 1 q k=1 q=1 p=1 j k ( 1)p 1[ q p ][ n+1 q k+2 p (q 1)!(n q)! ] = n j. (8)
23 first sum formula cont d The lefthand side of equation (8) is q q=1 p=1 If p = 1, the inner sum is ( 1) p 1[ ] q p (q 1)!(n q)! n 1 k=1 [ ] n + 1 q j k. k + 2 p n 1 [ ] n + 1 q n [ ] [ ] n + 1 q n + 1 q j k = j k 1 = k + 1 k 1 k=1 k=1 [ ] [( ) ] (n + j q)! n + 1 q n + j q = (n q)! 1, j(j 1)! 1 j where we have used the generating function (6).
24 first sum formula cont d If p 2, the inner sum is [ n + 1 q k k 1 ] j k+p 2 = j again using equation (6), and so k=1 p 2 (n + j q)!, (j 1)! n 1 [ ] ( ) n + 1 q n + j q j k = (n q)!j p 1. k + 2 p j Thus the lefthand side of equation (8) is ( ) n + j q j + j q=1 q q=2 p=2 ( 1) p 1[ q p (q 1)! ]( n + j q j ) j p 1.
25 first sum formula cont d The double sum can be rewritten as ( ) n + j q 1 j (q 1)! q=2 ( ) n + j q q=2 = j q=2 1 (q 1)! ( ) n + j q + j q p=2 [ ] q ( j) p 1 = p [ (1 j)(2 j) (q 1 j) ( )( ) n + j q j 1 j q 1 q=2 ( 1) q 1 [ ]] q 1 using equation (6) again, so the lefthand side of equation (8) is ( )( ) n + j q j 1 j + ( 1) q 1. j q 1 q=1
26 One more lemma This means we need one last lemma. Lemma For positive integers n and j, Proof. ( )( ) n + j q j 1 ( 1) q 1 = n. j q 1 q=1 Use Wilf s snake oil method : let F (x) = n=1 q=1 ( n + j q j )( j 1 q 1 ) ( 1) q 1 x n.
27 One more lemma cont d Proof cont d. Then F (x) = = q=1 q=1 ( j 1 q 1 = ( ) j 1 ( 1) q 1 q 1 n=1 ) ( 1) q 1 x q ( n + j q (1 x) j+1 = x (1 x) j+1 j ) x n j 1 ( j 1 p p=0 ) ( x) p x(1 x)j 1 (1 x) j+1 = x (1 x) 2 = x + 2x 2 + 3x 3 + and the result follows.
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