Chapter 9 Series III With the exceptio of the Null Sequece Test, all the tests for series covergece ad divergece that we have cosidered so far have dealt oly with series of oegative terms. Series with both positive ad egative terms are harder to deal with. 9. Alteratig Series Oe very special case is a series whose terms alterate i sig from positive to egative. That is, series of the form ( ) + a where a 0. Example ( ) + = + 3 4 + 5... is a alteratig series. Theorem Alteratig Series Test Suppose (a ) is decreasig ad ull. The the alteratig series ( ) + a is coverget. Proof. We first observe that the subsequece (s ) coverges: s = k= ( ) k+ a k = (a a )+(a 3 a 4 )+...+(a a ) = (a k a k ). This is a sum of oegative terms. It does ot ted to ifiity, because it is bouded: s = a (a a 3 ) (a 4 a 5 )... a a. The (s ) coverges to a umber l. There remais to show that the whole sequece (s ) coverges to l. For every ε > 0, there exists N such that for all eve m > N, s m l < ε. If m is odd, we have s m l s m+ l + s m+ s m < ε + a m+. This is less tha ε if m is large eough, sice the sequece (a m ) teds to 0. 85 k=
86 CHAPTER 9. SERIES III Example Sice ( ) is a decreasig ull sequece of this test tells us right away that ( ) + = + 3 4 + 5... is coverget. ( ) Similarly, is a decreasig ull sequece, therefore ( ) + = + 3 4 + 5... is coverget. Pairig terms i a suitable fashio, as i the proof above, oe ca get the followig error bouds. Exercise Show that, if (a ) is a decreasig ad ull sequece, the ( ) k+ a k an. k=n Exercise Lets = s 0 6. = ( ) +. FidavalueofN sothat N = ( ) + The Alteratig Series test requires that the sequece be decreasig ad ull, hece it must be o-egative. The ext exercise shows that if we relax either the decreasig or ull coditio the the alteratig series may ot coverge - eve if we still isist o the terms beig o-egative. Exercise 3 Fid a sequece (a ) which is o-egative ad decreasig but where ( ) + a is diverget ad a sequece (b ) which is o-egative ad ull but where ( ) + b is diverget. Exercise 4 The series 3! + 5! 7! + 9!... coverges to si. Explai how to use the series to calculate si to withi a error of 0 0. Exercise 5 Usig the Alteratig Series Test where appropriate, show that each of the followig series is coverget.. ( ) + 3 +. ( ) 3. cos π +( ) 4. π (+) 3 si 9. Geeral Series Series with positive terms are easier because we ca attempt to prove that the partial sums (s ) coverge by exploitig the fact that (s ) is icreasig. For a geeral series a, we get some iformatio by studyig the series of absolute values, a, which ivolves oly positive terms.
9.. GENERAL SERIES 87 Defiitio The series a is absolutely coverget if a is coverget. Example The alteratig series ( ) + is absolutely coverget because ( ) + = is coverget. The series ( ) is ot absolutely coverget because diverges. The series ( ) ( is absolutely coverget because ) coverges. Exercise 6 Is the series ( ) + absolutely coverget? Exercise 7 For what values of x is the Geometric Series x absolutely coverget? Absolutely coverget series are importat for the followig reaso. Theorem Absolute Covergece Every absolutely coverget series is coverget. Proof. Let s = i= a i ad t = i= a i. We kow that (t ) is coverget, hece Cauchy: for every ε > 0, there exists N such that t m t < ε for all m, > N. We ow show that (s ) is also Cauchy. Let > m. s s m = k=m+ a k k=m+ a k = t t m < ε. The (s ) is Cauchy, ad it coverges. Exercise 8 Is the coverse of the theorem true: Every coverget series is absolutely coverget? The Absolute Covergece Theorem breathes ew life ito all the tests we developed for series with o-egative terms: if we ca show that a is coverget, usigoeofthesetests, theweareguareteedthat a coverges as well. Exercise 9 Show that the series si is coverget. We see that 0 si. Therefore si is coverget by the Compariso Test. It follows that si is coverget by the Absolute Covergece Theorem. The Ratio Test ca be modified to cope directly with series of mixed terms. Theorem Ratio Test Suppose a 0 for all ad a+ a l. The a coverges absolutely (ad hece coverges) if 0 l < ad diverges if l >.
88 CHAPTER 9. SERIES III Proof. If 0 l <, the a coverges by the old Ratio Test. Therefore a coverges by the Absolute Covergece Theorem. If l >, we are guarateed that a diverges, but this does ot, i itself, prove that a diverges (why ot?). We have to go back ad modify our origial proof. We kow that there exists N such that a+ a whe > N. The a + a a... a N > 0. Therefore the sequece (a ) does ot ted to 0, ad a diverges by the ull sequece test. Example Cosider the series x. Whe x = 0 the series is coverget. (Notice that we caot use the Ratio Test i this case.) Now let a = x. Whe x 0 the a+ a = x+ + x = + x x. Therefore x is coverget whe x < ad diverget whe x >, by the Ratio Test. What if x =? Whe x = the x = which is diverget. Whe x = the x = ( )+ which is coverget. Theorem Ratio Test Variat Suppose a 0 for all ad a+ a, the a diverges. Exercise 0 Prove this theorem. Exercise I the ext questio you will eed to use the fact that if a o-egative sequece (a ) a ad a > 0, the ( a ) a. Prove this, by first showig that a a = a a a + a. Exercise Determie for which values of x the followig series coverge ad diverge. [Make sure you do t eglect those values for which the Ratio Test does t apply.]. x!. x 3.!x 4. (x) 5. (4x) 3 + 6. ( x)
9.3. EULER S CONSTANT 89 9.3 Euler s Costat Our last aim i this Chapter is to fid a explicit formula for the sum of the alteratig series: + 3 4 + 5 6 +... O the way we shall meet Euler s costat, usually deoted by γ, which occurs i several places i mathematics, especially i umber theory. Exercise 3 Let D = i= i + x dx = i= i log(+).. Usig Figure 9., draw the areas represeted by D.. Show that (D ) is icreasig. 3. Show that (D ) is bouded - ad hece coverget. f Euler s Costat The limit of the sequece D = i= i log(+) is called Euler s Costat ad is usually deoted by γ (gamma). Its value has bee computed to over 00 decimal places. To 4 decimal places, it is 0.57756649053. No-oe kows whether γ is ratioal or irratioal. f f3 f4 f- f 3 4 5 - + Figure 9.: Calculatig a upper boud of a itegral. ThelimitofthesequeceD = i= i log(+)iscalledeuler scostat ad is usually deoted by γ. Exercise 4 Show that i= ( ) i+ i evaluate ( ) +. Hit: First, derive the followig idetity: + 3 4 + + = + + 3 + + = log + D D. Hece ( + + 3 + ) 4 +...
90 CHAPTER 9. SERIES III 9.4 * Applicatio - Stirlig s Formula * Usig the alteratig series test we ca improve the approximatios to! that we stated i workbook 4. Take a look at what we did there: we obtaied upper ad lower bouds to log(!) by usig block approximatios to the itegral of logxdx. To get a better approximatio we use the approximatio i Figure 9.. a b a3 b3 a4 a5 b4 b5 a b b 3 4 5 Figure 9.: Approximatig the itegral by the mid poit. Now the area of the blocks approximates logxdx except that there are small triagular errors below the graph (marked as b,b,b 3,...) ad small triagular errors above the graph (marked as a,a 3,a 4,...). Note that log! = log+log3+ +log = area of the blocks. Exercise 5 Use the above diagram to show: log! ( + ) log+ = b +a b +a 3 b 3 + b +a [Hit: logxdx = log +] The curve logx is cocave ad it seems reasoable (ad ca be easily proved - try for yourselves), that a b a + ad (a ) 0. Exercise 6 Assumig that these claims are true, explai why (s ) = ( b +a b + +a b ) coverges. This proves that log! = (+ )log +Σ where Σ teds to a costat as. Takig expoetials we obtai:! costat e
9.4. * APPLICATION - STIRLING S FORMULA * 9 What is the costat? This was idetified with oly a little more work by the mathematicia James Stirlig. Ideed, he proved that: a result kow as Stirlig s formula.! e π as Check Your Progress By the ed of this chapter you should be able to: Use ad justify the Alteratig Series Test: If (a ) is a decreasig ull sequece the ( ) + a is coverget. Use the proof of Alteratig Series Test to establish error bouds. Say what it meas for a series to be absolutely coverget ad give examples of such series. Prove that a absolutely coverget series is coverget, but that the coverse is ot true. Use the modified Ratio Test to determie the covergece or divergece of series with positive ad egative terms. Prove that = ( ) + = log.
9 CHAPTER 9. SERIES III