16.2. Infinite Series. Introduction. Prerequisites. Learning Outcomes

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1 Infinite Series 6. Introduction We extend the concet of a finite series, met in section, to the situation in which the number of terms increase without bound. We define what is meant by an infinite series being convergent by considering the artial sums of the series. As rime examles of infinite series we examine the harmonic and the alternating harmonic series and show that the former is divergent and the latter is convergent. We consider various tests for the convergence of series, in articular we introduce the Ratio test which is a test alicable to series of ositive terms. Finally we define the meaning of the terms absolute and conditional convergence. Prerequisites Before starting this Section you should... Learning Outcomes After comleting this Section you should be able to... be able to use the summation notation be familiar with the roerties of limits 3 be able to use inequalities use the alternating series test and the ratio test on infinite series understand the terms absolute and conditional convergence

2 . Introduction Many of the series considered in section were examles of finite series in that they all involved the summation of a finite number of terms. When the number of terms in the series increases without bound we refer to the sum as an infinite series. Of articular concern with infinite series is whether they are convergent or divergent. For examle, the infinite series ++++ is clearly divergent because the sum of the first n terms increases without bound as more and more terms are taken. It is less clear as to whether the harmonic and alternating harmonic series: converge or diverge. Indeed you may be surrised to find that the first is divergent and the second is convergent. What we shall do in this section is to consider some simle convergence tests for infinite series. Although we all have an intuitive idea as to the meaning of convergence of an infinite series we must be more recise in our aroach. We need a definition for convergence which we can aly rigorously. First, using an obvious extension of the notation we have used for a finite sum of terms we denote the infinite series: a + a + a a + by the exression where a is an exression for the th term in the series. So, as examles: a ++3+ = = since the th term is a since the th term is a = ( ) + here a ( )+ Consider the infinite series: a + a + + a + = a What we do is to consider the sequence of artial sums, S,S,...,ofthis series where S = a S = a + a. S n = a + a + + a n HELM (VERSION : March 8, 004): Workbook Level 6.: Infinite Series

3 That is, S n is the sum of the first n terms of the infinite series. If the limit of the sequence S,S,...,S n,... can be found; that is lim S n = S n then we define the sum of the infinite series to be S: S = a (say) and we say the series converges to S. Another way of stating this is to say that a = lim n n a Definition An infinite series Convergence of Infinite Series a is convergent if the sequence of artial sums S,S,S 3,...,S k,... in which S k = k a is convergent Divergence condition for an infinite series An almost obvious requirement that an infinite series should be convergent is that the individual terms in the series should get smaller and smaller. This leads to the following keyoint: The condition: Key Point a 0 as increases (mathematically lim a =0) is a necessary condition for the convergence of the series a It is not ossible for an infinite series to be convergent unless this condition holds. 3 HELM (VERSION : March 8, 004): Workbook Level 6.: Infinite Series

4 Which of the following series cannot be convergent? (a) (b) (c) In each case, use the condition from the revious Keyoint. (a) a = lim a = a = lim + = + Hence series is divergent. (b) a = lim a = a = lim a =0 so this series may be convergent. Whether it is or not requires further testing. (c) a = lim a = a = ( )+ lim a =0so again this series may be convergent. Divergence of the harmonic series The harmonic series: has a general term a n = which clearly gets smaller and smaller as n. However, surrisingly, the series is divergent. Its divergence is demonstrated by showing that the harmonic n series is greater than a series which is obviously divergent. We do this by grouing the terms of the harmonic series in a articular way: ( ) ( 3 + ) ( ) + 8 HELM (VERSION : March 8, 004): Workbook Level 6.: Infinite Series 4

5 Now ( ) ( ) > = ( ) 6 > = > = etc and so on. Hence the harmonic series satisfies: ( ) ( ) ( ) + 8 ( ) > + + ( ) + ( ) + The right-hand side of this inequality is clearly divergent so the harmonic series is divergent Convergence of the alternating harmonic series As with the harmonic series we shall grou the terms of the alternating harmonic series, this time to dislay its convergence. The alternating harmonic series is: S = This series may be re-groued in two distinct ways. st re-grouing ( 5 + = ) ( 3 4 ) ( 5 6 ) 7 each term in brackets is ositive since >, > and so on. So we easily conclude that S< since we are subtracting only ositive numbers from. nd re-grouing ( 5 + = ) ( + 3 ) ( ) + 6 Again, each term in brackets is ositive since >, >, > and so on So we can also argue that S> since we are adding only ositive numbers to the value of the first term,. The conclusion that is forced uon us is that <S< so the alternating series is convergent since its sum, S, lies in the range in Section 6.5 that S =ln It will be shown 5 HELM (VERSION : March 8, 004): Workbook Level 6.: Infinite Series

6 . General Tests for Convergence The techniques we have alied to analyse the harmonic and the alternating harmonic series are one-off :- they cannot be alied to infinite series in general. However, there are many tests that can be used to determine the convergence roerties of infinite series. Of the large number available we shall only consider two such tests in detail. The alternating series test An alternating series is a secial tye of series in which the sign changes from one term to the next. They have the form a a + a 3 a 4 + (in which each a i,i=,, 3,... is a ositive number) Examles are: (a) + + (b) (c) For series of this tye there is a simle criterion for convergence: The alternating series Key Point The Alternating Series Test a a + a 3 a 4 + (in which each a i,i=,, 3,... are ositive numbers) is convergent if and only if the terms continually decrease: the terms decrease to zero: a >a >a 3 >... a 0 as increases (mathematically lim a =0) This is called the alternating series test. Which of the following series are convergent (a) ( ) ( ) ( +) (b) ( ) + HELM (VERSION : March 8, 004): Workbook Level 6.: Infinite Series 6

7 (a) First, write out the series: Now examine the series for convergence ( ) ( +) = ( ( + as increases. Since the individual terms of the series do not converge ) to zero this is therefore a divergent series. (b) Aly the rocedure used in (a) to roblem (b). This series is an alternating series of the form a a + a 3 a 4 + in which a =. The a sequence is a decreasing sequence since > > 3 >... Also lim ) =0. Hence the series is convergent by the alternating series test. 7 HELM (VERSION : March 8, 004): Workbook Level 6.: Infinite Series

8 3. The Ratio Test This test, which is one of the most useful and widely used convergence tests, alies only to series of ositive terms. Key Point The Ratio Test Let a beaseries of ositive terms. Suose, as increases, the limit of a + a a + anumber λ. That is lim = λ. Then, it is ossible to show that: a equals if λ>, then if λ<, then if λ =, then a diverges a converges a may be convergent or divergent. That is, the test is inconclusive in this case. Examle Use the ratio test to examine the convergence of the series (a) ! 3! 4! (b) +x + x + x 3 + Solution (i) The general term in this series is! i.e. +! + 3! + =! a =! a + = ( + )! and the ratio a + a =! ( + )! = ( )...(3)()() ( +)( )...(3)()() = ( +) HELM (VERSION : March 8, 004): Workbook Level 6.: Infinite Series 8

9 Solution (contd.) a + lim a = lim ( +) =0 Since 0 < the series is convergent. In fact, it will be easily shown using the techniques outlined in in Section 6.5 that +! + + =e.78 3! (ii) Here we must assume that x>0since we can only aly the ratio test to a series of ositive terms. Now +x + x + x 3 + = x so that and a + lim a a = x, a + = x x = lim = lim x x = x Thus, using the ratio test we deduce that (if x is a ositive number) this series will only converge if x<. We will see in Section 6.4 that +x + x + x 3 + = x rovided 0 <x<. (relace x by x and choose = inthe Binomial series). Use the ratio test to examine the convergence of the series: ln (ln 3) + 7 (ln 3) 3 + First, find the general term of the series. a = = ln 3 (ln 3) 3 so a (ln 3) = 3 (ln 3) Now find a + a + = 9 HELM (VERSION : March 8, 004): Workbook Level 6.: Infinite Series

10 a + = (+)3 (ln 3) + Finally, obtain lim a + a a + a + = lim = a a a+ a = ( + )3 (. Now + (ln 3) )3 (ln 3) < Hence this is a convergent series. = ( + )3 a + as increases lim = a Note that in all of these examles and guided exercises we have decided uon the convergence or divergence of various series; we have not been able to use the tests to discover what actual number the convergent series converges to. 4. Absolute and Conditional Convergence The ratio test alies to series of ositive terms. Indeed this is true of many related tests for convergence. However, as we have seen, not all series are series of ositive terms. To aly the ratio test such series must first be converted into series of ositive terms. This is easily done. Consider two series a and a. The latter series, obviously directly related to the first, is a series of ositive terms. Using imrecise language, it is harder for the second series to converge than it is for the first, since, in the first, some of the terms may be negative and cancel out art of the contribution from the ositive terms. No such cancellations can take lace in the second series since they are all ositive terms. Thus it is lausible that if a converges so does a. This leads to the following definition. Definition Conditional Convergence A convergent series a for which its related series conditionally convergent Absolute Convergence A convergent series a is said to be absolutely convergent if a is divergent is said to be a is convergent. HELM (VERSION : March 8, 004): Workbook Level 6.: Infinite Series 0

11 For examle, the alternating harmonic series: ( ) + = is conditionally convergent since the series of ositive terms ( ) + = is divergent. Show that the series! + 4! + is absolutely convergent. 6! First, find the general term of the series + + = ( ) a! 4! 6!! + 4! 6! + = ( ) ()! a ( ) ()! The related series of ositive terms is = ( ) a! 4! 6! = Now use the ratio test to examine the convergence of this series th term = ( +) th term = so a = ()! ()! [ ] ( +) th term What is lim? th term [ ] ( +) th term lim = th term th term = ( +) th term = ()! ((+))! HELM (VERSION : March 8, 004): Workbook Level 6.: Infinite Series

12 ()! (( + ))! = ( )... ( + )( + )( )... = ( + )( +) So the series of ositive terms is convergent by the ratio test. Hence convergent. 0as increases. ( ) ()! is absolutely Exercises. Which of the following alternating series are convergent? ( ) ln(3) ( ) + sin( +) π (a) (b) (c) + ( + 00). Use the ratio test to examine the convergence of the series: e 4 3 (a) (b) (c) ( +) +! (d) (0.3) (e) ( ) For what values of x are the following series absolutely convergent? ( ) x ( ) x (a) (b)! Answers. (a) convergent, (b) convergent, (c) divergent. (a) λ = 0so convergent, (b) λ = 0so convergent, (c) λ = so test is inconclusive. However, since > then the given series is divergent by comarison with the harmonic / series. (d) λ = 0/3 sodivergent, (e) Not a series of ositive terms so the ratio test cannot be alied. x 3. (a) The related series of ositive terms is. For this series, using the ratio test we find λ = x so the original series is absolutely convergent if x <. x (b) The related series of ositive terms is.for this series, using the ratio test we find! λ = 0(irresective of the value of x) sothe original series is absolutely convergent for all values of x. HELM (VERSION : March 8, 004): Workbook Level 6.: Infinite Series

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