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The Comparison Test— Not Just for Nonnegative Series

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The Comparison Test— Not Just for Nonnegative Series VOL. 79, NO. 3, JUNE 2006 205 13. TRUE. Given such a set of ellipsoidal balls, project the ellipsoidal balls orthogonally onto a line perpendicular to π. By Helly’s theorem in dimension one, these projec- tions must have a common point. Take a plane π parallel to π through this com- mon point. 14. FALSE. (1 − x)−1(1 − x 3)−1(1 − x 9)−1(1 − x 15)−1 ∞ ∞ ∞ ∞ = x a x 3b x 9c x 15d , a=0 b=0 c=0 d=0 so we need to count the number of 4-tuples (a, b, c, d) of nonnegative integers such that a + 3b + 9c + 15d = 45. First, a must be a multiple of 3, say a = 3A, which leads to A + b + 3c + 5d = 15. d = 3givesA = b = c = 0. (1 way.) d = 2givesc = 1 or 0 giving 3 and 6 possibilities for the pair (A, b).(9ways.) d = 1givesc = 3, 2, 1 or 0, giving 2, 5, 8, and 11 possibilities. (26 ways.) d = 0givesc = 5, 4, 3, 2, 1, or 0, giving 1, 4, 7, 10, 13, and 16 possibilities. (51 ways.) There are 87 ways, so the coefficient of x 45 is 87. Since the initial writing of this Note, the Bernoulli Trials were held in 2005 for the (n + 1)st time. After 9 rounds of competition, only 7 students remained standing, and of these only 1 had made no mistakes to that point. Round 10 saw the elimination of all but one of the remaining competitors, leaving the winner, Lino Demasi. But then ties needed to be broken so these 6 eliminated competitors were brought back in for Round 11, which eliminated all but two of them: Ian Baillargeon and Ralph Furmaniak. In a valiant effort to break this tie for 2nd/3rd place, the organizers tried to get them to crack under pressure. After two rounds with equal responses, a tie was declared for 2nd/3rd place. The Comparison Test— Not Just for Nonnegative Series MICHELE LONGO Universita` Cattolica del Sacro Cuore Milan, Italy [email protected] VINCENZO VALORI Universita` degli Studi di Firenze Florence, Italy [email protected]fi.it Is it possible to generalize the Comparison Test to generic real series? More precisely, ≤ ≤ is it true that, given an bn cn for all sufficiently large natural numbers n, the con- vergence of bn follows from the convergence of an and cn? At first glance, 206 MATHEMATICS MAGAZINE many of us (certainly the authors) could argue something like “If it were true then it would certainly appear in some of the books standing on the shelves in my office.” As a matter of fact, all the books on the authors’ shelves state the test only for nonneg- ative series. In particular, Hardy [5, p. 376, lines 1–3] considers possible extensions involving the comparison of only two series and affirms: “. there are no comparison tests for convergence of conditionally convergent series.” A generalization of the comparison test The Comparison Test is usually stated only for nonnegative real series, both in calculus books [1, 4, 5, 8]andinmorespe- cialized texts [2, 3, 6, 7]. There could be many reasons for this; however, the most immediate generalization can be used to establish the convergence or divergence of many series for which standard tests do not apply. Also, the proof is so straightforward that, at least, it could be considered as an exercise in calculus courses. The Comparison Test can be generalized as follows. THEOREM. Let an, bn, and cn be three real series such that an ≤ bn ≤ cn for all sufficiently large n ∈ N, then: (i) the series bn converges if an and cn converge; +∞ +∞ (ii) the series bn diverges to if an diverges to ; and (iii) the series bn diverges to −∞ if cn diverges to −∞. ≤ ≤ ≥ Proof. Assume an bn cn for all natural numbers n N.(i)If an and ( − ) ≤ − ≤ − cn converge, then cn an converges. The relation 0 bn an cn an ( − ) = and the usual Comparison Test imply that bn an converges. Since bn (bn − an) + an, we conclude that bn converges. Parts (ii) and (iii) follow from the relation M M M an ≤ bn ≤ cn, n=N n=N n=N which is true for each M ≥ N. In the Theorem, the only requirement on an and cn is that they converge. Clearly, the case of interest is that of conditionally convergent series (that is, conver- gent but not absolutely convergent, see Hardy [5, p. 375]). It is worth noting that the ≤ proof of our Theorem can be adapted to show that if an bn and one of the two series an or bn converges then the other must either converge or else have an infinite limit. We give now an example. EXAMPLE 1. Consider the alternating real series bn where the generic term (−1)n b = ln 1 + n nγ depends on the positive real parameter γ . Normally, we would apply the Alternating Series Test (also known as Leibniz Test) which states that an alternating series bn converges if |bn| is monotonic decreasing and bn → 0[3, p. 55]. However, the test applies in this case if and only if γ ≥ 1, because otherwise the sequence |bn| is not decreasing. To see this, observe that, if n is even, n (−1) 1 2 ln 1 + = ln 1 + = ln 1 + ; nγ nγ 2nγ − 1 + (−1)n VOL. 79, NO. 3, JUNE 2006 207 whereas, if n is odd, 1 1 2 |b | =−ln 1 − = ln 1 + = ln 1 + . n nγ nγ − 1 2nγ − 1 + (−1)n γ n Therefore, |bn| = ln (1 + 2/ (2n − 1 + (−1) )) for every n ∈ N and γ γ n+1 |bn| ≥ |bn+1| ⇔ n ≤ (n + 1) + (−1) . The second inequality is verified for all positive γ if n is odd, but holds only for γ ≥ 1ifn is even. Furthermore, bn is absolutely convergent if and only if γ>1, as follows. The power series expansion of ln(1 + x) allows us to assert that |x| /2 ≤ |ln (1 + x)| ≤ 2 |x|, as long as |x| is sufficiently small; this in turn implies that 0 < 1/2nγ < ln (1 + (−1)n /nγ ) < 2/nγ for sufficiently large n and the usual Comparison Test yields the result. This analysis shows that we cannot use standard tests to study bn when 0 <γ < 1, so let us proceed as follows: First observe that x − 3x 2/4 ≤ ln (1 + x) ≤ x − x 2/4 in a neighborhood of zero (again from the power series expansion), so the substitution x = (−1)n /nγ yields (−1)n 3 (−1)n (−1)n 1 − ≤ ln 1 + ≤ − nγ 4n2γ nγ nγ 4n2γ n γ for sufficiently large n. Second, notice that by the Alternating Series Test, (−1) /n converges for all γ>0 whereas 1/n2γ converges if and only if γ>1/2. Finally, n γ 2γ n γ 2γ by setting an = (−1) /n − 3/4n and cn = (−1) /n − 1/4n , an application of our Theorem enables us to conclude that 1 1 if γ> , then b converges, and if 0 <γ ≤ , then b diverges to −∞. 2 n 2 n The previous example suggests a general procedure that can be useful to decompose the general term of the series into simpler quantities, which can be analyzed separately. Our generalized Comparison Test then applies to give information about the original series. This procedure is summarized in the following proposition. PROPOSITION. Let an be convergent and suppose f is a real valued function such that, in a neighborhood of 0, f (x) = αx + βx 2k + o x 2k ,β= 0, k ∈ N. 2k Then f (an) converges if and only if (an) converges. Proof. By assumption there exists ε>0 such that for |x| <εwe have |β| |β| αx + β − x 2k ≤ f (x) ≤ αx + β + x 2k . 2 2 Hence, there exists Nε such that, for each n > Nε, |β| |β| αa + β − (a )2k ≤ f (a ) ≤ αa + β + (a )2k . n 2 n n n 2 n 208 MATHEMATICS MAGAZINE ( ) ( )2k Now, convergence of f an follows from convergence of an and part (i)of ( )2k the Theorem. Conversely, if an diverges then parts (ii)and(iii) imply that also f (an) diverges. REMARK. The Proposition cannot be extended to the case where the expansion ends with an odd power (Examples 4 and 5 show this point). Nevertheless, if an converges and f (x) = αx + βx 2k+1 + o x 2k+1 ,β= 0, k ∈ N, then 2k+1 |an| converges ⇒ f (an) converges. (1) | |2k+1 Finally, notice that a sufficient condition for the convergence of an is the con- 2i vergence of (an) for some i ∈ {1, 2,...,k}. This Remark gives only a sufficient condition for the convergence of f (an). For this reason, it may be possible to obtain more information about the convergence of the series by refining the Taylor expansion of f . For example, consider the se- √ √ 2k+1 ries arctan (−1)n / 4 n . In this case (−1)n / 4 n converges for all nat- ural numbers k whereas it converges absolutely if and only if k ≥ 2; so if we use arctan x = x − x 3/3 + o x 3 , we can say nothing about the convergence of the series √ since (−1)n / 4 n3 does not converge; on the contrary, if we consider the expan- sion up to the fifth order we can conclude that the series is convergent. A less trivial example of this occurrence is the following. EXAMPLE 2. Consider the alternating real series bn, whose generic term, de- pending on the positive real parameter γ ,isdefinedby (−1)n 1 b = tan + .
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