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3.2 Introduction to Infinite Series

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3.2 Introduction to Infinite Series 3.2 Introduction to Infinite Series Many of our infinite sequences, for the remainder of the course, will be defined by sums. For example, the sequence m X 1 S := : (1) m 2n n=1 is defined by a sum. Its terms (partial sums) are 1 ; 2 1 1 3 + = ; 2 4 4 1 1 1 7 + + = ; 2 4 8 8 1 1 1 1 15 + + + = ; 2 4 8 16 16 ::: These infinite sequences defined by sums are called infinite series. Review of sigma notation The Greek letter Σ used in this notation indicates that we are adding (\summing") elements of a certain pattern. (We used this notation back in Calculus 1, when we first looked at integrals.) Here our sums may be “infinite”; when this occurs, we are really looking at a limit. Resources An introduction to sequences a standard part of single variable calculus. It is covered in every calculus textbook. For example, one might look at * section 11.3 (Integral test), 11.4, (Comparison tests) , 11.5 (Ratio & Root tests), 11.6 (Alternating, abs. conv & cond. conv) in Calculus, Early Transcendentals (11th ed., 2006) by Thomas, Weir, Hass, Giordano (Pearson) * section 11.3 (Integral test), 11.4, (comparison tests), 11.5 (alternating series), 11.6, (Absolute conv, ratio and root), 11.7 (summary) in Calculus, Early Transcendentals (6th ed., 2008) by Stewart (Cengage) * sections 8.3 (Integral), 8.4 (Comparison), 8.5 (alternating), 8.6, Absolute conv, ratio and root, in Calculus, Early Transcendentals (1st ed., 2011) by Tan (Cengage) Integral tests, comparison tests, ratio & root tests. * section 9.4 (Convergence Tests), 9.5 (Comparison, ratio, root tests), 9.6 (Alternating, abs. conv & cond. conv) Calculus, Early Transcendentals (11th ed., 2009) by Anton, Bivens, Davis (John Wiley & Sons) p. 645 of Anton has a nice list. * section 10.3 (integral test), 10.4 (alt series), 10.5 (comparison), 10.6 (absolute convergence), 10.7 (ra- tio and root) in the Whitman College online textbook: http://www.whitman.edu/mathematics/ multivariable/ * Whitman's online textbook: http://www.whitman.edu/mathematics/multivariable/calculus_ 10_Sequences_and_Series.pdf 1 3.2.1 What is a series? 1 m 1 X X X 1 Given an infinite series a we define the partial sum S := a : Thus, in the series we have n m n 2n n=1 n=1 n=1 the partial sums 1 S = 1 2 3 S = 2 4 7 S = 3 8 15 S = 4 16 1 X We mean, by the expression, , the limit, as n ! 1, of the partial sums Sn: n=1 1 In this case, the partial sums appear to have the pattern S = 1 − : So n 2n 1 X 1 2n n=1 really means m X 1 1 lim = lim 1 − = 1: m!1 2n m!1 2m n=1 Since, in this case, the limit is 1, we say that 1 X 1 = 1: 2n n=1 3.2.2 An easy divergence test Intuitively, if a series is to converge to a finite limit L we would expect that eventually the terms we are adding up are contributing very little to the series { that at some point the sum is close to L and each new term is not changing that. This argument can be made precise (but we won't do that here.) This gives us a theorem, the \n-th term divergence test": Theorem. (n-th term test) X If a series an converges then the limit, as n goes to infinity, of the terms an, must be zero. Although stated in terms of convergence, the theorem is really a statement about divergence, for it is equivalent to the statement: X The n-th term divergence test: If in a series an the terms an do not go to zero then the series does not converge! (This statement is the contrapositive of the statement in the theorem.) 3.2.3 Geometric series The complexity of our investigation into series (initiated long ago by the Bernoulli brothers and Euler) can be displayed by examining the family of geometric series { central to our understanding of series { and two interesting \sporadic" series, the harmonic series and the alternating harmonic series. 2 In an earlier section, we examined the series 1 X 1 1 1 1 1 = + + + + ::: 1 + :::: 2n 2 4 8 16 2n n=1 1 We concluded, just by observation, that the partial sums had the form 1 − 2n and thus the series converged to 1. 1 Note that this series has this property: each term added on is exactly 2 of the previous term. A series which has a \ratio" r such that each new term is exactly r times the previous term, is said to be geometric. (See http://en.wikipedia.org/wiki/Geometric_series for a general discussion of these series, including modern applications.) The main idea. There is a nice way to work out a formula for the partial sum of a geometric series. In general, a geometric series has form 1 X a + ar + ar2 + ar3 + ::: + arn−1 + ::: = arn n=0 where a is the first term and r is the common ratio between terms. Let us write m−1 2 3 m−1 X n Sm := a + ar + ar + ar + ::: + ar = ar : n=0 Notice that m 2 3 m−1 m X n rSm := ar + ar + ar + ::: + ar + ar = ar n=1 and so m rSm − Sm := ar − a: (Notice how most terms cancel!) Therefore a(rm − 1) a(1 − rm) S = = : m r − 1 1 − r a(1 − rm) If jrj > 1 then the expression does not converge and so the geometric series does not 1 − r converge. a(1−rm) If r = 1 then the expression 1−r is undefined but it is easy to check that the partial sums are Sm = ma and so the series diverges to infinity. a(1−rm) If r = −1 then the 1−r does not converge and so the geometric series does not converge. The partial sums alternate between a and 0: a(1−rm) a a But if jrj < 1 then the 1−r converges to 1−r : Therefore the geometric series converges to 1−r : This is important enough to emphasize as a theorem. Theorem. (Geometric series) If jrj < 1 then 1 X a a + ar + ar2 + ar3 + ::: + arn−1 + ::: = arn = : 1 − r n=0 1 X But if jrj ≥ 1 then arn diverges. n=0 3 The geometric series are central to the study of infinite series. We will see later in this course that if a series is not geometric, we will attempt (in a certain way) to \pretend" it is geometric anyway! Sometimes this \pretense" gives us very useful information. (This will motivate the \ratio" and \root" tests.) 3.2.4 The Harmonic Series The n-th term divergence test says that the terms of a series must go to zero if there is any hope of the series converging. Warning! It is tempting to believe in the converse statement. Is it true that if the terms go to zero then the series converges? That would be nice { but it is not true in general. Here is an example { a classical one { where the terms go to zero but the series diverge. The series is called the \harmonic series". The harmonic series 1 X 1 1 1 1 1 = 1 + + + + ::: + + :::: n 2 3 4 n n (See http://en.wikipedia.org/wiki/Harmonic_series_(mathematics) for Wikipedia's discussion of this series, including an explanation for its name.) 1 The terms go to zero as n goes to infinity. Yet this series diverges! n We give one argument that this series diverges. This argument is a \comparison test". Another argument will be given later. Compare the series 1 X 1 1 1 1 1 1 1 1 1 1 1 = 1 + + + + + + + + + + ::: + + ::: n 2 3 4 5 6 7 8 9 10 n n and the series 1 1 1 1 1 1 1 1 1 1 1 + + ( + ) + ( + + + ) + ( + + ::: ) + ::: 2 4 4 8 8 8 8 16 16 16 1 In the second series, any expression of the form , 2s < n ≤ 2s+1, we replace n by 2s+1: Since n 1 1 ≤ , we have a series which is \smaller" than the harmonic series. Each partial sum of the second 2s+1 n series no bigger than the partial sum of the harmonic series. 1 But notice that, since in the second series there are 2s terms of the form then we can collect 2s+1 1 them together to form the term : 2 So the second series becomes 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 + + ( + ) + ( + + + ) + ( + + ::: ) + ::: = 1 + + ( ) + ( ) + ( ) + ::: + ( ) + ::: 2 4 4 8 8 8 8 16 16 16 2 2 2 2 2 | {z } 8 terms 1 We continue, forever, to add 2 in the sum. So this second series diverges (by the \n-th term test", if you will.) But since the harmonic series is forever larger than this diverging series, then the harmonic series diverges! The harmonic series forms a nice counterpoint to the n-th term test for divergence. If the n-th term of a series goes to zero, the series still might not converge. The harmonic series is a nice example of this phenomenon. 4 3.2.5 The Alternating Harmonic Series One more interesting series: Consider the alternating harmonic series 1 X (−1)n+1 1 1 1 1 (−1)n+1 = 1 − + − − + ::: + + :::: n 2 3 4 5 n n Note the effect of the expression (−1)n+1; it forces the signs to alternate, so that we are adding positive, then negative terms.
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