Abel Summation MOP 2007, Black Group

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1 Abel Summation MOP 007, Blac Group Zachary Abel June 5, 007 This lecture focuses on the Abel Summation formula, which is most often useful as a way to tae advantage of unusual given conditions such as ordering or majorization, or simply as a way to put a new loo on an epression. But in addition to learning to use this formula, I want to emphasize good, motivated thining 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. Abel Summation Let n be a positive integer, and let a, a,..., a n and b, b,..., b n be two sequences. Then if S = a + +a, = n a b = S n b n + S b b + ) ) = a )b b ) + a + a )b b 3 ) + + a + + a n )b n b n ) + a + + a n )b n ). = This is often called Summation by Parts due to its continuous analog, Integration by Parts: f) g ) d = f) g) f ) g) d. ) Here, differentiation corresponds to differencing and antidifferentiation corresponds to partial sums.) Indeed, the equivalence of ) and ) can be established directly: one direction with Riemann sums, and the other with appropriate choice of functions. Problem. Do this. The following problems can eplain the usefulness of this formula better than I could while waving my hands, so let s get started. Problem. Given two sets of real numbers a, a,..., a n ; b, b,..., b n, prove that the following two statements are equivalent: i). For any numbers, a + a + + a n b + b + + b n.

2 ii). a + + a n = b + + b n and a + + a b + + b for each n. Proof. Suppose the first condition is true. Setting i = for each i gives a i b i, and similarly setting = for each i gives a i b i, so the sums are equal. Net, for each, setting = = =, + = = = 0 shows that a + + a b + + b for each, which is eactly the condition in ii). Now suppose ii) holds, and tae any sequence. Define A = a + + a and B = b + + b. By Abel Summation, we have a + + a n = A ) + A 3 ) + + A n ) + A n B ) + B 3 ) + + A n ) + B n 3) = b + + b n, where line 3) follows because i i+ 0 and A n = B n. This shows i), and we re done. Problem 3 Romania 999). a). Let, y,, y,...,, y n be positive real numbers such that i). y < y < < y n ; ii) y + y + + y for all =,,..., n. Prove that y y y n b). Let A = {a, a,..., a n } N be a set such that for all distinct subsets B, C A, B that <. a a a n Proof. a). The difference of the two sides can be epressed as i= y i i = i= i y i i y i. By Abel Summation, we can manipulate this to mae use of the given conditions: i= i y i i y i = y n y ) + + y n ) ) + 0, n i= i y i i+ y i+ C. Prove ) y ) + + i y i ) ) as needed. b). Order the numbers a < a < < a n. The nonempty subsets of {a,..., a } have all different sums, so a + + a for each. If we set y =, this reads a + + a y + + y.

3 Also, we clearly have a y < a y < < a n y n since both sequences strictly increase. Thus, by part a), we have as desired = <, a a n y y n n Problem 4. Define the sequence u,..., u n by u = and u n = u u n + for n =, 3,.... Prove that for all n, the closest under-approimation of by n Egyptian fractions is u u n. I.e., if < < < are distinct positive integers such that + +, then in fact ) u u n Proof. First, notice that u + + u n = u u n. Since + + can be epressed as a fraction with denominator at most, if < u u n, then inequality 4) must hold. So suppose that u u n. The proof is by induction. The base case is clear, as u =. Now, we d lie to prove + + u + + u n. By hypothesis we now that + + u + + u n, so if we new that u n, the result would follow. Unfortunately, we don t now much about the size of, so this fails. What if we pull off the terms l, l+,..., instead of just, for some l? If we could show that l + u l + u n, the result would follow by adding this to the inductive hypothesis applied to inde l. The same problem that the i s may be much smaller than the u i s might occur, but this would seem to go against the assumption that u u u n. This idea is what saves us. Let l be the largest inde j so that j j+ u j u j+ u n. This implies l u l, l l+ u l u l+, l l+ l+ u l u l+ u l+, etc... because otherwise we could increase l. It turns out that these inequalities alone are enough to prove that , l l+ u l u l+ u n which lets us finish by induction as mentioned. We ll show this fact in the following form which is equivalent to the original by taing reciprocals): Lemma. If p p p n and q q q n are two decreasing sequences of positive numbers and if p p p q q q for each n, then p + + p n q + + q n. 3

4 We can use Abel Summation in the following way why would we thin to use Abel Summation?): q + + q n = = = = p q p p p + ) = q + + q ) p p q q p p + ) p p p p + ) = = p, where we use the convention p n+ = 0. This is the desired inequality. Problem 5. For any real number and positive integer n, prove that = = sin π. Proof. After playing at the series for a while, one of the difficulties that seems to arise is the eistence of two types of terms: the early ones where sin influences the sum more than the denominator, and the later ones where the taes over. So, a natural idea is to split the sum into two parts, namely m = sin and =m+ sin, for some inde 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 < π in fact 0 < < π, as the = 0 case is easy), we can use the inequality sin t t to perhaps somewhat stupidly) form the bound m = sin m = = m. 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 lie to do something simple lie simply say sin t and then let the s in the denominator carry the rest. While this doesn t quite wor in its current form the harmonic series diverges), we can use Abel Summation to rewrite the sum as follows: =m+ sin = n S n + n =m+ nn + ) S, 5) 4

5 where S = sinm + )) + + sin). But notice that we can mae each S telescope as follows: S = sint) = t=m+ t=m+ sin cos t + ) ) ) cos t = sin ) ) cos n + cos m + ). Now we can use the sin t bound: we find S sin =m+ sin sin n + n =m+ = sin, and so 5) produces the following: ) = + m + ) sin. So we have the bound m + m+) sin on the given sum, where m is some integer that has not yet been chosen. As this epression is probably minimized when both terms are equal, perhaps we should try to show each term is bounded by π π, i.e. we should try choosing m =. Then m + > π, so we would need sin π. But this is indeed true, as the line y = π connecting 0, sin 0) = 0, 0) with π, sin π ) = π, ) lies below the curve y = sin on the interval [0, π ] since sin is concave down here). Problems Problem 6 Abel s Inequality). Let n be a positive integer, and let a, a,..., a n and b, b,..., b n be two sequences of real numbers with b b n 0. Then if S = a + +a, m = min we have mb a i b i Mb. i= S, and M = ma S, Problem 7. Calculate the values of n = and n = using both Abel Summation and differentiation of geometric series. Compare to the previous discussion of Abel Summation and Integration by Parts. Which derivation was quicer? Problem 8. Let a, b, c, d 0 be such that a, a + b 5, a + b + c 4, and a + b + c + d 30. Show that a + b + c + d 0. Problem 9 98 USAMO). If is a positive real number and n is a positive integer, then prove that [n] [] + [] + [3] [n] n, where [t] denotes the greatest integer less than or equal to t. Problem 0 USA TST 007 #). Let n be a positive integer and let a a a n and b b b n be two non-decreasing sequences of real numbers such that a + + a i b + + b i for every i =,..., n and a + + a n = b + + b n. Suppose that for any real number m, the number of pairs i, j) with a i a j = m equals the number of pairs, l) with b b l = m. Prove that a i = b i for i =,..., n. 5

6 Problem. Consider a polygonal line P 0 P P n such that P 0 P P = P P P 3 = = P n P n P n, with all angles measured counterclocwise. If P 0 P > P P > > P n P n, show that P 0 and P n cannot coincide. Problem Romanian Selection 007, aa Blue Homewor). Let a, a,..., a n be real numbers such that a i for all i, and a + a + + a n = 0. Prove that there eists n such that Can this bound be improved? a + a + + a +. 4 Problem 3 MOP 006). Let,,..., and y, y,..., y n be real numbers with < < < < and < y < y < < y n. Given that y y for every integer with n, prove that ) ) ) ) ) y y ). y n Problem 4 Oleg Golberg, MOP 006). Given real numbers a, a,..., a n, where n is an integer greater than, prove that there eist real numbers b, b,..., b n satisfying the following conditions: a). a i b i are positive integers for i n; and b). b i b j ) n. i<j n 6