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Abel Summation MOP 2007, Black Group Abel Summation MOP 2007, Black Group Zachary Abel June 25, 2007 This lecture focuses on the Abel Summation formula, which is most often useful as a way to take advantage of unusual given conditions such as ordering or majorization, or simply as a way to put a new look on an expression. But in addition to learning to use this formula, I want to emphasize good, motivated thinking for all of these problems. Don’t simply hunt for the time and place to apply Abel Summation; hopefully, when and if the occasion arises, you’ll recognize it. 1 Abel Summation Let n be a positive integer, and let a1, a2, . , an and b1, b2, . , bn be two sequences. Then if Sk = a1+···+ak, n n−1 X X akbk = Snbn + Sk(bk − bk+1) (1) k=1 k=1 = (a1)(b1 − b2) + (a1 + a2)(b2 − b3) + ······ + (a1 + ··· + an−1)(bn−1 − bn) + (a1 + ··· + an)(bn). This is often called “Summation by Parts” due to its continuous analog, Integration by Parts: Z Z f(x) · g0(x) dx = f(x) · g(x) − f 0(x) · g(x) dx. (2) (Here, differentiation corresponds to differencing and antidifferentiation corresponds to partial sums.) Indeed, the equivalence of (1) and (2) can be established directly: one direction with Riemann sums, and the other with appropriate choice of functions. Problem 1. Do this. The following problems can explain the usefulness of this formula better than I could while waving my hands, so let’s get started. Problem 2. Given two sets of real numbers a1, a2, . , an; b1, b2, . , bn, prove that the following two state- ments are equivalent: (i). For any numbers x1 ≤ x2 ≤ · · · ≤ xn, a1x1 + a2x2 + ··· + anxn ≤ b1x1 + b2x2 + ··· + bnxn. 1 (ii). a1 + ··· + an = b1 + ··· + bn and a1 + ··· + ak ≥ b1 + ··· + bk for each 1 ≤ k ≤ n. P P Proof. Suppose the first condition is true. Setting xi = 1 for each i gives ai ≤ bi, and similarly setting P P x= − 1 for each i gives ai ≥ bi, so the sums are equal. Next, for each k, setting x1 = ··· = xk = −1, xk+1 = ··· = xn = 0 shows that a1 + ··· + ak ≥ b1 + ··· + bk for each k, which is exactly the condition in (ii). Now suppose (ii) holds, and take any sequence x1 ≤ · · · ≤ xn. Define Ak = a1 + ··· + ak and Bk = b1 + ··· + bk. By Abel Summation, we have a1x1 + ··· + anxn = A1(x1 − x2) + A2(x2 − x3) + ··· + An−1(xn−1 − xn) + Anxn ≥ B1(x1 − x2) + B2(x2 − x3) + ··· + An−1(xn−1 − xn) + Bnxn (3) = b1x1 + ··· + bnxn, where line (3) follows because xi − xi+1 ≤ 0 and An = Bn. This shows (i), and we’re done. Problem 3 (Romania 1999). (a). Let x1, y1, x2, y2, . , xn, yn be positive real numbers such that (i). x1y1 < x2y2 < ··· < xnyn; (ii). x1 + x2 + ··· + xk ≥ y1 + y2 + ··· + yk for all k = 1, 2, . , n. Prove that 1 1 1 1 1 1 + + ··· + ≤ + + ··· + x1 x2 xn y1 y2 yn X X (b). Let A = {a1, a2, . , an} ⊆ N be a set such that for all distinct subsets B, C ⊆ A, x 6= x. Prove x∈B x∈C that 1 1 1 + + ··· + < 2. a1 a2 an Proof. (a). The difference of the two sides can be expressed as n n X 1 1 X xi − yi − = . y x x y i=1 i i i=1 i i By Abel Summation, we can manipulate this to make use of the given conditions: n X xi − yi x y i=1 i i n−1 1 X 1 1 = (x − y ) + ··· + (x − y ) + − (x − y ) + ··· + (x − y ) x y 1 1 n n x y x y 1 1 i i n n i=1 i i i+1 i+1 ≥ 0, as needed. k (b). Order the numbers a1 < a2 < ··· < an. The 2 − 1 nonempty subsets of {a1, . , ak} have all different k k−1 sums, so a1 + ··· + ak ≥ 2 − 1 for each k. If we set yk = 2 , this reads a1 + ··· + ak ≥ y1 + ··· + yk. 2 Also, we clearly have a1y1 < a2y2 < ··· < anyn since both sequences strictly increase. Thus, by part (a), we have 1 1 1 1 1 + ··· + ≤ + ··· + = 2 − n−1 < 2, a1 an y1 yn 2 as desired. Problem 4. Define the sequence u1, . , un by u1 = 2 and un = u1 ··· un−1 + 1 for n = 2, 3,.... Prove that 1 for all n, the closest under-approximation of 1 by n Egyptian fractions is 1− . I.e., if x1 < x2 < ··· < xn u1···un are distinct positive integers such that 1 + ··· + 1 ≤ 1, then in fact x1 xn 1 1 1 1 + ··· + ≤ + ··· + . (4) x1 xn u1 un Proof. First, notice that 1 + ··· + 1 = 1 − 1 . Since 1 + ··· + 1 can be expressed as a fraction u1 un u1···un x1 xn with denominator at most x1 ··· xn, if x1 ··· xn < u1 ··· un, then inequality (4) must hold. So suppose that x1 ··· xn ≥ u1 ··· un. 1 1 The proof is by induction. The base case is clear, as x1 ≥ u1 = 2. Now, we’d like to prove +···+ ≤ x1 xn 1 1 1 1 1 1 + ··· + . By hypothesis we know that + ··· + ≤ + ··· + , so if we knew that xn ≥ un, u1 un x1 xn−1 u1 un−1 the result would follow. Unfortunately, we don’t know much about the size of xn, so this fails. What if we pull off the terms x`, x`+1, . , xn instead of just xn, for some `? If we could show that 1 + ··· 1 ≤ 1 + ··· 1 , the result would follow by adding this to the inductive hypothesis applied to index x` xn u` un ` − 1. The same problem that the xi’s may be much smaller than the ui’s might occur, but this would seem to go against the assumption that x1 · x2 ··· xn ≥ u1 · u2 ··· un. This idea is what saves us. Let ` be the largest index j so that xj · xj+1 ··· xn ≥ uj · uj+1 ··· un. This implies x` ≥ u`, x`x`+1 ≥ u`u`+1, x`x`+1x`+2 ≥ u`u`+1u`+2, etc. because otherwise we could increase `. It turns out that these inequalities alone are enough to prove that 1 1 1 1 1 1 + + ··· + ≤ + + ··· + , x` x`+1 xn u` u`+1 un which lets us finish by induction as mentioned. We’ll show this fact in the following form (which is equivalent to the original by taking reciprocals): Lemma 1. If p1 ≥ p2 ≥ · · · ≥ pn and q1 ≥ q2 ≥ · · · ≥ qn are two decreasing sequences of positive numbers and if p1 · p2 ··· pk ≤ q1 · q2 ··· qk for each 1 ≤ k ≤ n, then p1 + ··· + pn ≤ q1 + ··· + qn. 3 We can use Abel Summation in the following way (why would we think to use Abel Summation?): n X qk q1 + ··· + qn = pk · pk k=1 n X q1 qk = (pk − pk+1) + ··· + p1 pk k=1 n r q ··· q X k 1 k ≥ (pk − pk+1) · k p1 ··· pk k=1 n X ≥ (pk − pk+1) · k k=1 n X = pk, k=1 where we use the convention pn+1 = 0. This is the desired inequality. Problem 5. For any real number x and positive integer n, prove that n X sin kx √ ≤ 2 π. k k=1 Proof. After playing at the series for a while, one of the difficulties that seems to arise is the existence of two types of terms: the early ones where sin kx influences the sum more than the denominator, and the later 1 ones where the k takes over. So, a natural idea is to split the sum into two parts, namely m n X sin kx X sin kx and , k k k=1 k=m+1 for some index m yet to be determined. Consider the first of these sums. These are the ones where the sin terms seem to dominate. So assuming WLOG that 0 ≤ x < π (in fact 0 < x < π, as the x = 0 case is easy), we can use the inequality sin t ≤ t to (perhaps somewhat stupidly) form the bound m m X sin kx X kx ≤ = mx. k x k=1 k=1 Let’s be a bit more clever for the second of the sums. Recall that these are the terms where we consider the denominators more influential, so we’d like to do something simple like simply say sin t ≤ 1 and then let the k’s in the denominator carry the rest. While this doesn’t quite work in its current form (the harmonic series diverges), we can use Abel Summation to rewrite the sum as follows: n n−1 X sin kx 1 X 1 = S + S , (5) x n n n(n + 1) k k=m+1 k=m+1 4 where Sk = sin((m + 1)x) + ··· + sin(kx). But notice that we can make each Sk telescope as follows: k k X X 1 1 S = sin(tx) = cos t + 1 x − cos t − 1 x = cos n + 1 x − cos m + 1 x . k sin x 2 2 2 sin x 2 2 t=m+1 t=m+1 2 2 2 1 Now we can use the | sin t| ≤ 1 bound: we find |Sk| ≤ x = x , and so (5) produces the following: 2 sin 2 sin 2 n n−1 ! X sin kx 1 1 X 1 1 1 ≤ · + − = . k sin x n k k + 1 (m + 1) sin x k=m+1 2 k=m+1 2 1 So we have the bound mx + x on the given sum, where m is some integer that has not yet been (m+1) sin 2 chosen. As this expression is probably minimized when both terms are equal, perhaps we should try to show √ √ √ j π k π each term is bounded by π, i.e.
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