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. we should try choosing m = x . Then m + 1 > x , so we would need x x 2x π π π sin 2 ≥ π . But this is indeed true, as the line y = π connecting (0, sin 0) = (0, 0) with ( 2 , sin 2 ) = ( 2 , 1) π lies below the curve y = sin x on the interval [0, 2 ] (since sin is concave down here).

2 Problems

Problem 6 (Abel’s Inequality). Let n be a positive integer, and let a1, a2, . . . , an and b1, b2, . . . , bn be two

sequences of real numbers with b1 ≥ · · · ≥ bn ≥ 0. Then if Sk = a1 +···+ak, m = min Sk, and M = max Sk, k k we have n X mb1 ≤ aibi ≤ Mb1. i=1 Pn k Pn 2 k Problem 7. Calculate the values of k=1 kx and k=1 k x using both Abel Summation and diﬀerenti- ation of geometric series. Compare to the previous discussion of Abel Summation and Integration by Parts. Which derivation was quicker?

Problem 8. Let a, b, c, d ≥ 0 be such that a ≤ 1, a + b ≤ 5, a + b + c ≤ 14, and a + b + c + d ≤ 30. Show √ √ √ √ that a + b + c + d ≤ 10.

Problem 9 (1982 USAMO). If x is a positive real number and n is a positive integer, then prove that

[x] [2x] [3x] [nx] [nx] ≥ + + + ··· + , 1 2 3 n

where [t] denotes the greatest integer less than or equal to t.

Problem 10 (USA TST 2007 #2). Let n be a positive integer and let a1 ≤ a2 ≤ · · · ≤ an and b1 ≤ b2 ≤

· · · ≤ bn be two non-decreasing sequences of real numbers such that

a1 + ··· + ai ≤ b1 + ··· + bi for every i = 1, . . . , n − 1 and

a1 + ··· + an = b1 + ··· + bn.

Suppose that for any real number m, the number of pairs (i, j) with ai − aj = m equals the number of pairs

(k, `) with bk − b` = m. Prove that ai = bi for i = 1, . . . , n.

5 Problem 11. Consider a polygonal line P0P1 ··· Pn such that ∠P0P1P2 = ∠P1P2P3 = ··· = ∠Pn−2Pn−1Pn, with all angles measured counterclockwise. If P0P1 > P1P2 > ··· > Pn−1Pn, show that P0 and Pn cannot coincide.

Problem 12 (Romanian Selection 2007, aka Blue Homework). Let a1, a2, . . . , an be real numbers such that

|ai| ≤ 1 for all i, and a1 + a2 + ··· + an = 0. Prove that there exists k ≤ n such that

2k + 1 |a + 2a + ··· + ka | ≤ . 1 2 k 4

Can this bound be improved?

Problem 13 (MOP 2006). Let x1, x2, . . . , xn and y1, y2, . . . , yn be real numbers with 1 < x1 < x2 < ··· < xn

and 1 < y1 < y2 < ··· < yn. Given that x1 ··· xk ≥ y1 ··· yk for every integer k with 1 ≤ k ≤ n, prove that

1 1 1 1 1 1 1 − 1 − ··· 1 − ≥ 1 − 1 − ··· 1 − . x1 x2 xn y1 y2 yn

Problem 14 (Oleg Golberg, MOP 2006). Given real numbers a1, a2, . . . , an, where n is an integer greater

than 1, prove that there exist real numbers b1, b2, . . . , bn satisfying the following conditions:

(a). ai − bi are positive integers for 1 ≤ i ≤ n; and X n2 − 1 (b). (b − b )2 ≤ . i j 12 1≤i