MAT026 Calculus II Basic Covergece Tests for Series Egi MERMUT 202.03.08 Dokuz Eylül Uiversity Faculty of Sciece Departmet of Mathematics İzmir/TURKEY Cotets Mootoe Covergece Theorem 2 2 Series of Real Numbers 2 3 Series of Noegative Real Numbers 4 4 Itegral Test ad p-series 5 5 Compariso ad Limit Compariso Tests 6 6 Root ad Ratio Tests 7 7 Alteratig Series Test 2 8 Coditioally Coverget Series like Alteratig Harmoic Series 5 9 Absolutely Coverget Series are Coverget 5 0 Rearragemet Theorems for Absolutely ad Coditioally Coverget Series 7 Basic Tests for Covergece of a Series 8 2 Iterval of Covergece of Power Series 8 See Your Lecture Notes ad Textbook for Proofs SEE THE LECTURE NOTES AND YOUR TEXTBOOK FOR THE PROOF OF THE FOL- LOWING THEOREMS FOR BASIC CONVERGENCE TESTS FOR SERIES OF REAL NUMBERS.
Mootoe Covergece Theorem Defiitio of coverget sequeces of real umbers A sequece of real umbers is a real-valued fuctio whose domai is the set of positive itegers. It is deoted by (a ) where a is a real umber for each Z +. For a sequece (a ) of real umbers, we defie:. The sequece (a ) is said to coverge to a real umber L if for every positive real umber ε, there is a idex N Z + such that for all Z +, > N a L < ε. I this case, we write a L as, or lim a = L. 2. The sequece (a ) is said to be coverget if there exists a real umber L such that (a ) coverges to L. Mootoe Covergece Theorem Completeess of the real umber system R (the least upper boud property of R) is also equivalet to the followig property: Theorem. Mootoe Covergece Theorem. A mootoe sequece (a ) umbers is coverget if ad oly if it is bouded. More precisely: of real. If (a ) is a odecreasig sequece bouded from above, the it coverges to L = sup({a Z + }). 2. If (a ). is a odecreasig sequece that is ot bouded from above, the it diverges to 3. If (a ) is a oicreasig sequece bouded from below, the it coverges to L = if({a Z + }). 4. If (a ). is a oicreasig sequece that is ot bouded from below, the it diverges to 2 Series of Real Numbers Defiitio of coverget series ad diverget series For a sequece (a ) of real umbers, let s = a k = a + a 2 + + a for all Z +. k= The sequece (s ) is called the sequece of partial sums of the series a, ad a is called the th term of the series a. We say that the series a is coverget if the 2
sequece (s ) of partial sums is coverget. If (s ) coverges to a real umber S, we write a = S ad say that the series a has sum S: a is diverget if the sequece (s ) of partial sums is diverget. We say that the series Geometric Series [ ] a = S = lim s = lim a k. k= Theorem 2. For a real umber r, the geometric series r = + r + r 2 + r 3 + r 4 + + r + =0 is coverget if ad oly if r <. If r <, the the geometric series Harmoic Series If a series =0 =0 r = r r has sum r : is diverget ad lim = 0 if r <. a of real umbers is coverget, the we must have lim a = 0. See the lecture otes for its proof. Never misuse this if-the statemet. It does ot say that if lim a = 0, the the series a is coverget which is a wrog statemet. For a series a, we may have lim a = 0 but the series a may be diverget. For example, if a = for all Z+, the we have the so called Harmoic Series a = which is a diverget series but lim a = lim = 0. We have give i the lectures two proofs for the divergece of the Harmoic Series. Divergece Test If a series a of real umbers is coverget, the we must have lim a = 0 as we said above. The cotrapositive of this statemet gives us: 3
Theorem 3. Divergece Test. Let (a ) be a sequece of real umbers. If diverget. lim a does ot exist or if lim a exists but lim a 0, the the series a is The covergece or divergece of a series is ot affected if we chage fiitely may of its terms. So i the tests for covergece, the hypothesis for the terms of the series eeds to hold for all large idexes, that is, for all idexes after a fixed idex. For example, if we say that 0 a b for all large Z +, the we mea that there exists a idex N 0 Z + such that 0 a b holds for all itegers > N 0. Liear Combiatios of Coverget Series Theorem 4. Let a ad b be coverget series of real umbers with sums A ad B: a = A ad b = B The for every real umber c, all of the followig series are coverget (a + b ), (a b ), ca ad have the followig sums: (a + b ) = A + B, (a b ) = A B, ca = ca. See the lecture otes for the proof of this theorem, ad for the proof of the followig corollary: Corollary 5. If ay two of the followig three series is coverget, the the third oe is also coverget: a, b, (a + b ). 3 Series of Noegative Real Numbers Series of oegative terms The compariso tests, itegral test, ratio ad root tests are tests for series with oegative terms. This is a very importat hypothesis because oly uder this case we have the followig: Theorem 6. Series of oegative terms. Let a be a series of real umbers such that The series a 0 for all Z +. a is coverget if ad oly if there exists a real umber M such that a k M for all Z +. k= 4
The proof of this theorem is obtaied by applyig the Mootoe Covergece Theorem to the sequece (s ) of partial sums of the series a. Proof of the theorem for series of oegative terms Let a be a series of real umbers such that a 0 for all Z +. Let s = a k for all Z +. The the sequece (s ) of partial sums of the series a is a odecreasig sequece because for all Z +, [ + ] s + = a k = a k + a + = s + a + s sice a + 0. k= k= So by the Mootoe Covergece Theorem, the sequece (s ) of partial sums is coverget if ad oly if it is bouded from above. This gives us that the series a is coverget if ad oly if there exists a real umber M such that s = k= a k M for all Z +. 4 Itegral Test ad p-series Itegral Test Theorem 7. Itegral Test. Suppose that the fuctio f : [, ) R is a positive decreasig ad cotiuous fuctio o [, ). Let The the series a = f() for all Z +. a is coverget if ad oly if the improper itegral If the improper itegral + for the remaider R = S s. f(x) dx coverges, the we have the followig estimate f(x) dx R = S s f(x) dx for all Z +. k= f(x) dx is coverget. I the above theorem, our fuctio f(x) may be defied o [N, ) for some positive iteger N (istead of the iterval [, )); similar result the holds for the series a ad the improper itegral N f(x) dx. See your lecture otes for the proof of the itegral test. The proof is give by comparig areas uder the graph of y = f(x) ad the sum of the areas of rectagles approximatig below ad above. 5 =N
p-series is coverget if ad oly if p > p Theorem 8. For a real umber p R, the p-series is coverget if ad oly if p >. p Prove this theorem. If p 0, the it is diverget by Divergece Test. If p > 0, the use the Itegral Test by checkig that it satisfies all the hypothesis of the itegral test. Remember that the improper itegral dx is coverget if ad oly if p >. xp 5 Compariso ad Limit Compariso Tests Compariso Test Theorem 9. Compariso Test. Let (a ) ad (b ) be two sequeces of real umbers such that 0 a b for all large Z +, that is, there exists N 0 Z + such that. If 2. If b coverges, the a diverges, the Limit Compariso Test 0 a b for all itegers > N 0, a coverges. b diverges. Theorem 0. Limit Compariso Test. Let (a ) ad (b ) be two sequeces of real umbers such that a 0 ad b > 0 for all large Z +. a a Suppose that lim exists as a exteded real umber, that is, lim = L for some L b b [0, ) { }.. If 0 < L <, the 2. If L = 0 ad 3. If L = ad a coverges if ad oly if b coverges, the b diverges, the b coverges. a coverges. a diverges. 6
How to use Compariso Test? See the lecture otes for the proofs of the Compariso Test ad Limit Compariso Test. Usig Compariso Tests is ot easy ad you must try eough examples to lear how to make meaigful comparisos to decide whether a give series a is coverget or diverget. You must kow some coverget series ad diverget series that you ca use i your comparisos. For example, p-series p ad geometric series r are series whose behavior is well kow. Work o the problems i your textbook to get used to make comparisos ad usig compariso tests. Ifiite Decimal Expasios The ifiite decimal expasio 0.a a 2 a 3 a is defied to be the sum of the coverget series a 0 : 0.a a 2 a 3 a = a 0. where here a s are digits, that is, a {0,, 2, 3, 4, 5, 6, 7, 8, 9} a The series is coverget by Compariso Test because: 0. 0 a 0 9 0 for all Z+, ad 2. The geometric series 9 0 is coverget sice r = r = 0 <. For example, show that accordig to this defiitio 0.9 = 0.9999999 = 9 0 =. Try also some examples from your textbook ad lecture otes. Remember also how you were workig with examples for these periodic decimal expasios i your high school to prove that they are ratioal umbers. For example fid the ratioal umber whose decimal expasio is 92.38765 = 92.38765765765765765. Express it as a series of real umbers. Why does the method you have memorized i high school works? 6 Root ad Ratio Tests The Root Test ad Ratio Tests are proved by usig Compariso Test, comparig them with a geometric series, i the case whe the required limit is <. Divergece i the case whe the required limit is > follows by the Divergece Test. See the below detailed proofs of Root Test ad Ratio Test. 7
Root Test Theorem. Let (a ) be a sequece of real umbers with Suppose that lim a 0 for all Z +. a exists as a exteded real umber, say L = lim a for some L [0, ) { }.. If L <, the the series 2. If L >, the the series a is coverget. a is diverget. 3. If L =, the the test is icoclusive. Proof of Root Test i the case lim a = L <. Suppose lim a = L <. The there exists a real umber r > 0 such that L < r <, for example take r = (L + )/2. Show these o the real lie. Sice lim a = L, for the positive real umber ε = r L, there exists N Z + such that for all Z +, > N = a L < ε = r L = L ε < a < L + ε = r. Sice also a 0 for all Z +, we obtai: 0 a < r for all itegers > N. So we have 0 a < r for all itegers > N. Now use the Compariso Test for series of oegative real umbers. r is coverget sice r = r <. By the Compariso Test, the series a is coverget because we have:. 0 a < r for all itegers > N, ad 2. The series r is coverget. The geometric series Proof of Root Test i the case lim a = L >. Suppose lim a = L > ad L R. Let ε = L. Sice L >, ε = L > 0. Sice a = L, for the positive real umber ε = L, there exists N Z + such that for all lim Z +, > N = a L < ε = L = = L ε < a < L + ε. 8
Show this o the real lie. So we have: < a for all itegers > N. Show that such a N exists i the case L = also by usig the defiitio of lim Thus < a for all itegers > N. a =. The the sequece (a ) caot coverge to 0. This the implies by the Divergece Test that the series a is diverget. Proof of Root Test i the case lim Whe L = lim examples show. a =, the series a = L =. a may be coverget or diverget as the followig. Let a = for all Z+. The the harmoic series L = lim a = lim 2. Let a = 2 for all Z+. The the series p-series with p = 2 >, ad p L = lim a = lim a = / = lim a = / 2 = lim = =. is diverget, ad is coverget sice it is a 2 ( ) 2 = 2 =. So i this case, the root test is icoclusive ad you must try some other methods to determie covergece. Ratio Test Theorem 2. Let (a ) be a sequece of real umbers with a > 0 for all Z +. a + Suppose that lim exists as a exteded real umber, say a a + L = lim for some L [0, ) { }.. If L <, the the series 2. If L >, the the series a a is coverget. a is diverget. 3. If L =, the the test is icoclusive. 9
Proof of Ratio Test i the case lim Suppose lim a + a + a = L <. a = L <. The there exists a real umber r > 0 such that L < r <, for example take r = (L + )/2. Show these o the real lie. Sice lim positive real umber ε = r L, there exists N Z + such that for all Z +, > N = a + a Sice also a > 0 for all Z +, we obtai: So we have a + a L < ε = r L = L ε < a + < L + ε = r. a 0 < a + a < r for all itegers > N. 0 < a + < ra for all itegers > N. This gives for = N +, N + 2, N + 3,..., N + k,...: = N + = a N+2 < ra N+ = N + 2 = a N+3 < ra N+2 < r 2 a N+ = N + 3 = a N+4 < ra N+3 < r 3 a N+. =. = N + k + = 0 < a N+k+ < r k a N+ for all k Z + by iductio. = L, for the Now use the Compariso Test for series of oegative real umbers. The geometric series r k a N+ = a N+ r k is coverget sice r = r < (here a N+ is a fixed costat real k= k= umber, the idex for the series is k). By the Compariso Test, the series a N+k+ is coverget because we have: k=. 0 < a N+k+ < r k a N+ for all k Z +, ad 2. The series Sice the series r k a N+ is coverget. k= a N+k+ is coverget, we obtai that the series k= have just added N + terms to the begiig of the series). a + a is coverget (we Proof of Ratio Test i the case lim = L >. a a + Suppose lim = L > ad L R. Let ε = L. Sice L >, ε = L > 0. Sice a a + lim = L, for the positive real umber ε = L, there exists N Z + such that for all a Z +, > N = a + L a < ε = L = = L ε < a + < L + ε. a 0
Show this o the real lie. So we have: < a + a for all itegers > N. a + Show that such a N exists i the case L = also by usig the defiitio of lim =. a Sice also a > 0 for all Z +, we obtai: a + > a > 0 for all itegers > N. This gives for = N +, N + 2, N + 3,..., N + k,...: ad so a N+k+ > a N+k > > a N+3 > a N+2 > a N+ > 0, a N+k+ > a N+ > 0 for all k Z +. So the sequece (a N+k+ ) k= is a icreasig sequece of positive real umbers ad hece it caot coverge to 0. The the sequece (a ) caot coverge to 0. This the implies by the Divergece Test that the series a is diverget. a + Proof of Ratio Test i the case lim = L =. a a + Whe L = lim =, the series a may be coverget or diverget as the followig examples show. a. Let a = for all Z+. The the harmoic series a + L = lim a a = /( + ) = lim = lim / + =. is diverget, ad 2. Let a = 2 for all Z+. The the series p-series with p = 2 >, ad p a = is coverget sice it is a 2 a + L = lim a /( + ) 2 2 = lim / 2 = lim ( + ) 2 =. So i this case, the ratio test is icoclusive ad you must try some other methods to determie covergece. Root Test is stroger tha Ratio Test Theorem 3. Let (a ) be a sequece of real umbers with a > 0 for all Z +.
a + Suppose that lim a exists as a exteded real umber, say a + L = lim a for some L [0, ) { }. The the limit lim a also exists as a exteded real umber, ad equals L, that is, lim a + a = L = lim. a The proof of this theorem is left to your Aalysis courses. This theorem gives us that wheever Ratio Test works, so does Root Test. Hece Root Test is stroger tha Ratio Test. But i some cases you shall prefer Ratio Test ad i some cases you shall prefer Root Test; use the oe whose required limit seems easier to you. But kow that if Root Test fails, there is o eed to try Ratio Test by this theorem. 7 Alteratig Series Test Alteratig Series A series i which the terms are alterately positive ad egative is said to be a alteratig series, that is, a series a is said to be a alteratig series if for a sequece (u ) of positive real umbers either or so either or a = a = a = ( ) + u for all Z +, a = ( ) u for all Z + ; ( ) + u = u u 2 + u 3 u 4 + u 5 u 6 + u 7 u 8 + ( ) u = u + u 2 u 3 + u 4 u 5 + u 6 u 7 + u 8 + where u = a > 0 for all Z +. Alteratig Series Test (Leibiz s Theorem) Theorem 4. Suppose that (u ) three coditios: is a sequece of real umbers that satisfies the followig. u > 0 for all Z +, 2. There exists N 0 Z + such that 3. lim u = 0. u u + for all itegers N 0, 2
The the alteratig series ( ) + u is coverget. Say it has sum S = ( ) + u. Let s = k= ( )k+ u k for all Z +. For all itegers N 0, the remaider S s satisfies: S s u + ad sig(s s ) = ( ) (+)+ = ( ). That is we have the followig estimate for the remaider S s = S ( ) k+ u k. k= The absolute value of S s is less tha u + which is the absolute value of the first uused term ( ) (+)+ u +, ad the sig of S s is the sig of the first uused term ( ) (+)+ u +, that is, its sig is ( ) +2 = ( ). So we have 0 ( ) (S s ) u + for all itegers N 0. Proof of the Alteratig Series Test Without loss of geerality, let s assume that N 0 =, that is, u u + for all Z +. Let (s ) be the sequece of partial sums of the alteratig series ( ) + u : s = ( ) k+ u k for all Z +. k= Cosider the subsequeces (s 2m ) m= ad (s 2m ) m= of the sequece (s ) of partial sums. For these subsequeces, usig u u + ad u > 0 for all Z +, show by iductio that: (s 2m ) m= is a odecreasig sequece, that is, s 2m s 2(m+) for all m Z +. (s 2m ) m= is a oicreasig sequece, that is, s 2m s 2(m+) for all m Z +. s 2m < s 2m for all m Z + sice s 2m s 2m = u 2m < 0. s 2m < s 2 for all positive itegers m ad. Thus we obtai that s 2m s for all m Z +, ad so the odecreasig sequece (s 2m ) m= is bouded from above by s. Hece by the Mootoe Covergece Theorem, (s 2m ) m= is a coverget sequece, say with limit S: S = lim m s 2m. Similarly the oicreasig sequece (s 2m ) m= is bouded below by s 2. Hece by the Mootoe Covergece Theorem, (s 2m ) m= is a coverget sequece, say with limit S : S = lim m s 2m. 3
Sice s 2m s 2m = u 2m ad lim u = 0, we obtai S = lim m s 2m = lim m [s 2m + u 2m ] = S + 0 = S. Show all these o the real lie: s 2 s 4 s 6 lim m s 2m = S = lim m s 2m s 5 s 3 s. Hece the sequeces (s 2m ) m= ad (s 2m ) m= both coverge to the same real umber S. Sice these sequeces cosists of the terms of (s ) with eve idex ad odd idex, we obtai that the sequece (s ) of partial sums of the alteratig series ( ) + u coverges to S. This meas that the alteratig series ( ) + u has sum S: ( ) + u = lim s = S = lim s 2m = lim m m s 2m. Alteratig Harmoic Series The Alteratig Harmoic Series ( ) + is coverget, by the Alteratig Series Test, but the Harmoic Series sice it is a p-series series. Let a = ( )+ p ( ) + = is diverget with p =. Such series are called coditioally coverget for all Z +. The series a = ( ) + is coverget but a = is diverget. Before applyig the alteratig series that be sure that the give series is a alteratig series, that is, its terms are alterately positive ad egative. The check that it satisfies all of the above three coditios i the Alteratig Series Test to be able to use Alteratig Series Test to coclude that it is coverget. si Example 5. The series has ifiitely may positive ad egative terms but it is ot 2 a alteratig series, because its terms are ot alterately positive ad egative. For example the first three terms are positive, ad the the ext three terms are egative, etc. 4
8 Coditioally Coverget Series like Alteratig Harmoic Series Coditioally Coverget Series A series a is said to be a coditioally coverget series if. It is a coverget series, that is, 2. The series a is diverget. a is coverget, ad ( ) + The Alteratig Harmoic Series is a coditioally coverget series as we have show above sice: ( ) + The series is coverget ad ( ) + = series So if a series is diverget. a is a coditioally coverget series, the it is a coverget series but the a obtaied by takig absolute value of every term i the series 9 Absolutely Coverget Series are Coverget Absolutely Coverget Series A series a is said to be absolutely coverget if the series a is diverget. a coverges. A series a is said to be coditioally coverget if the series is coverget, but ot absolutely coverget. Theorem 6. Absolute Covergece Covergece. For a series if a is coverget, the the it is coverget. a is coverget. That is, if Proof of Absolute Covergece Covergece. a of real umbers, a is absolutely coverget, 5
series There is a ice trick to prove this theorem easily. Sice the series a is coverget, the 2 a is also coverget. Sice for every real umber x, we have x x x ad so 0 x + x 2 x, we obtai The both of the series 0 a + a 2 a for all Z +. (a + a ) ad we ca use Compariso Test for these series to obtai that the series because: 2 a are series of oegative real umbers. So 0 a + a 2 a for all Z +, ad 2 a is a coverget series. Sice ow both of the series is coverget, that is, the series series (a + a ) ad (a + a ) is coverget a are coverget, the series [(a + a ) a ] a is coverget. So whe you are give a series a to test for covergece, you must firstly look at the a by takig absolute values of the terms of the give series. Sice a 0 for all Z +, you ca apply the Compariso Test, Limit Compariso Test, Itegral Test, Ratio ad Root Tests of the previous sectios to the series a of oegative terms. If you fid that the series a is coverget, the sice absolute covergece implies covergece, you will obtai that the series a is coverget, ad moreover absolutely coverget. ( ) + The Alteratig Harmoic Series is coverget by the Alteratig Series Test but it is ot absolutely coverget sice the Harmoic Series ( ) + = is diverget. ( ) + So the Alteratig Harmoic Series is coditioally coverget. 6
si Example 7. The series is absolutely coverget because the series is coverget by the Compariso Test sice we have: 0 si 2 2 for all Z+, ad The series 2 is coverget. 2 Example 8. Show that for p R, the alteratig p-series is absolutely coverget if p >, ad coditioally coverget if 0 < p. ( ) + p si 2 0 Rearragemet Theorems for Absolutely ad Coditioally Coverget Series A coverget series is either absolutely or coditioally coverget Suppose that a is a coverget series. The either the series a is coverget or ot. So a coverget series a is either absolutely coverget or coditioally coverget. The behavior of these two kids of coverget series is very differet by the followig rearragemet theorems. A rearragemet of a series a is a series b formed by rearragig the terms of the series a, that is, there exists a oe-to-oe ad oto fuctio σ : Z + Z + such that b = a σ() for all Z +. Rearragemet Theorem for absolutely coverget series Theorem 9. If a series a is absolutely coverget ad if b is a rearragemet of this absolutely coverget series a, the the series has the same sum with the absolutely coverget series b = a. b is also absolutely coverget ad a : 7
Hece absolutely coverget series behave more like fiite sums with respect to rearragemets. So for example if a series a of positive real umbers is coverget (ad so automatically it is absolutely coverget as a > 0 for all Z + ), the every rearragemet coverget with the same sum: b = a. Rearragemet Theorem for coditioally coverget series Coditioally coverget series have a surprisig property ulike fiite sums: Theorem 20. If a series b is also a is coditioally coverget with sum S, the give ay real umber L, there exists a rearragemet such that the series b of this coditioally coverget series b has sum L. So a coditioally coverget series ca be rearraged to make the sum whatever you wish. Hece coditioally coverget series behave very differetly from absolutely coverget series with respect to rearragemets. This remarkable result shows how ifiite summatio of series is really differet tha fiite sums ad also shows that the order of summig i a series is really very importat. You shall see the proofs of these rearragemet theorems i your Aalysis courses. Basic Tests for Covergece of a Series See the two pages summary from Thomas Calculus review cards for the basic tests for covergece of series ad the steps that you must follow for testig a series for covergece for the examples that we shall work o. Solve the questios from your textbook give i Homework 2. To be able to use these tests for covergece, you must solve eough examples to become familiar with these tests ad to decide which test you must use. 2 Iterval of Covergece of Power Series See your lecture otes ad textbook for: Power series Examples for fidig the iterval of covergece of a give power series usig the above tests for covergece. Power Series A power series about x = 0 is a series of the form c x = c 0 + c x + c 2 x 2 + c 3 x 3 + + c x +, 8 a
where (c ) is a fixed sequece of real umbers. It is like a polyomial of ifiite degree. Our first questio for power series is to determie for which real umbers x the power series coverges. Istead of 0, we ca take ay real umber a: A power series about x = a is a series of the form c (x a) = c 0 + c (x a) + c 2 (x a) 2 + c 3 (x a) 3 + + c (x a) +, where a is a fixed real umber ad (c ) is a fixed sequece of real umbers. This power series is said to have ceter a ad the c s are said to be the coefficiets of this power series (like coefficiets of a polyomial). Iterval of covergece of a power series Give a power series c (x a), our first questio is For which real umbers x, this power series coverges? The set of all x R such that the power series c (x a) coverges is called the iterval of covergece of the power series. It is called a iterval because as you shall see i the below examples ad as we shall prove i geeral, this set is always a iterval i R with the ceter of iterval beig a; it may be of the followig types: {a}, (a R, a + R), [a R, a + R), (a R, a + R], [a R, a + R], (, ), for some R R. Such a R is called the radius of covergece of this power series. For the sigleto set {a} case, we take R = 0 ad for the (, ) case, we take R =. Radius of covergece of a power series If a power series c (x a) has radius of covergece R ad 0 < R, the we shall prove that the power series coverges absolutely for all x R such that x a < R, that is for all x i the ope iterval (a R, a + R). If 0 < R <, we shall show that the power series diverges for all x R such that x a > R; ad at the ed poits a R ad a + R, the power series may coverge at exactly oe of them or may coverge at both of them or may diverge at both of them, (ad whe it coverges at a ed poit, the covergece may be absolute covergece or coditioal covergece) so that the iterval of covergece will be ay of the followig four types whe 0 < R < : (a R, a + R), [a R, a + R), (a R, a + R], [a R, a + R]. Fidig the iterval of covergece of a power series Give a power series c (x a), to fid for which real umbers x it is absolutely coverget, apply for example Ratio Test or Root Test or some other tests for series of oegative real umbers to the series c (x a). There are oly three possible cases: 9
. The power series coverges absolutely oly at x = a ad diverges at all x a. I this case, the radius of covergece is take to be R = 0 ad the iterval of covergece is just the set [a, a] = {a} with oe elemet a. 2. The power series coverges absolutely for all x R. I this case, the radius of covergece is take to be R = ad the iterval of covergece is (, ) = (a, a + ). 3. There exists a real umber R > 0 such that the power series coverges absolutely o the ope iterval (a R, a+r) ad it diverges o R\[a R, a+r] = (, a R) (a+r, ). This R is called the radius of covergece of the power series. The to fid the iterval of covergece, you shall also determie if it coverges at the ed poits a R or a + R; it may be that it coverges at exactly oe of them or may coverge at both of them or may diverge at both of them, (ad whe it coverges at a ed poit, the covergece may be absolute covergece or coditioal covergece) so that the iterval of covergece may be ay of the followig four types: (a R, a + R), [a R, a + R), (a R, a + R], [a R, a + R]. Whe testig for covergece at the ed poits a R ad a + R, if the power series is ot absolutely coverget at a ed poit, the try Alteratig Series Test for the ed poit if possible. Of course, the basic tests are ot eough to fid the aswer for may series ad you shall lear some advaced tests for covergece i your Aalysis courses. Examples for fidig the iterval of covergece of a power series For each of the followig power series, fid the iterval of covergece ad the radius of covergece usig the basic covergece tests. Determie for which real umbers x, it is absolutely coverget ad for which real umbers x, it is coditioally coverget.. 2. 3. 4. ( ) ( ) =0 x!!x =0 x x2 2 See your textbook for the aswers to these problems. 20
Refereces [] Course textbook: Joel Hass, Maurice D. Weir, ad George B. Thomas, Jr. Uiversity Calculus: Elemets with Early Trascedetals, Pearso, 2009. [2] Spivak, Michael. Calculus. Corrected 3rd editio. Cambridge Uiversity Press, 2006. [3] Wade, William R. Itroductio to Aalysis. 4th editio, Pearso, 200. [4] Fitzpatrick, P. M. Advaced Calculus. Secod Revised editio. America Mathematical Society, 2009. 2