d) If the sequence of partial sums converges to a limit L, we say that the series converges and its

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1 Ifiite Series. Defiitios & covergece Defiitio... Let {a } be a sequece of real umbers. a) A expressio of the form a + a a +... is called a ifiite series. b) The umber a is called as the th term of the series. c) The sequece {s }, defied by s a k, is called the sequece of partial sums of the series. k d) If the sequece of partial sums coverges to a limit L, we say that the series coverges ad its sum is L. e) If the sequece of partial sums does ot coverge, we say that the series diverges. Examples... ) If 0 < x <, the x coverges to 0 x. Solutio. Let us cosider the sequece of partial sums {s }, where s x k. Here s x k x+ x x x+ x, N. k As, 0 < x <, x + 0 as. Hece s. Thus the give series coverges to x x. /// ) The series diverges. Solutio. Cosider the sequece of partial sums {s }, where s the subsequece s of {s }. Here s + / 3/, k k s 4 + / + /3 + /4 > 3/ + /4 + /4. k. Now, let us examie

2 Suppose s > ( + )/, the s + s + > + + k + + k + k + ( + ) + + Thus the subsequece {s } is ot bouded above ad as it is also icreasig, it diverges. Hece the sequece diverges, i.e., the series diverges. /// 3) (Telescopic series:) Show that the series coverges to. ( + ) Solutio. Cosider the sequece of partial sums {s }. The s k(k + ) k k k ( k ) k + +. Summarizig this observatio, oe has the followig theorem o Telescopic series Theorem..3. Suppose {a } is a sequece of o-egative real umbers such that a L. The the series (a a + ) coverges to a L. Lemma..4. ) If a coverges to L ad b coverges to M, the the series L + M. ) If a coverges to L ad if c R, the the series ca coverges to cl. Lemma..5. If a coverges, the lim a 0.. (a + b ) coverges to Proof. Suppose a L. The the sequece of partial sums {s } also coverges to L. Now a s s L L 0. /// Example..6. If x >, the the series x diverges. Solutio. Assume to the cotrary that the series x coverges. The the th term, i.e., x 0. But as x >, x for all N ad hece lim x, which is a cotradictio. Hece the series x diverges. ///

3 As a first result we have the followig compariso theorem: Theorem..7. Let {a }, {b } be sequeces of positive reals such that a b. If b coverges the a coverges. Proof. Let s a + a a ad t b + b b be the partial sum of a, b respectively. The s t. Sice b coverges, we have {t } coverges ad is bouded. Now sice {s } is mootoically icreasig sequece that is bouded above, we get the covergece of {s } ad hece the covergece of a. /// Theorem..8. Let {a } be a decreasig sequece of positive umbers. The a coverges if ad oly if a coverges. 0 Proof. Let s ad t be the sequece of partial sums of a ad a respectively. The s ad t are mootoically icreasig sequeces. We kow that such sequeces coverge if they are bouded from above. proof follows from the observatio that s k a a + a + (a 3 + a 4 ) + (a 5 + a 6 + a 7 + a 8 ) (a a ) a + a + a 4 + 4a 8 + 8a a a + t. (.) Therefore, if {s } coverges the {s } coverges ad hece bouded from above. Now covergece of {t } follows from., {t }. O the other had, s a + (a + a 3 ) + (a 4 + a 5 + a 6 + a 7 ) + (a a 5 ) + (a a ) a + a + 4a 4 + 8a a a + t So if {t } coverges, the {s } coverges. Now the coclusio follows from s s + ad the fact that {s } is mootoically icreasig sequece. /// Examples..9. ) Cosider the series p, p > 0. The, we have ( ) p ( which coverges for ) p p > ad diverges for p. ) Cosider the series log. Here give series diverges. log which diverges. log Hece the 3

4 . Absolute covergece Defiitio... b) If coverges absolutely. a coverges but a) Let a be a series of real umbers. If a diverges, we say that Examples... ) The series ( ) coverges absolutely.! a coverges, we say that a coverges coditioally. a ) The series ( ) coverges absolutely. Theorem..3. If a coverges absolutely, the a coverges. Proof. Let t a k. As the series coverges absolutely, the sequece {t } k ɛ > 0, there exists N N such that is Cauchy. Thus, give Let m >. The Thus the sequece {s } s m s t m t < ɛ m, N. m i+ a i m i+ a i t m t < ɛ. is Cauchy ad hece coverges. Thus a coverges. /// Theorem..4. Let a be a series of real umbers. Let p max{a, 0} ad q mi{a, 0}. a) If a coverges absolutely, the both p ad q coverges. b) If a diverges the oe of the p or q diverges. b) If a coverges coditioally the both p ad q diverges. Proof. a) Observe that p (a + a )/ ad q (a a )/. Thus the covergece of the two series follows from the hypothesis. b) Proof is easy. c) We leave to this as a exercise. /// 4

5 Tests for absolute covergece Theorem..5 (Compariso test). Let a be a series of real umbers. The, a coverges absolutely if there is a absolutely coverget series c with a c for all N, N N. Proof follows as i Theorem..7 Examples..6. ) The series 7 7 diverges because 7 7 /7 for all N ad diverges. ) The series! coverges because! ad coverges. 0 Theorem..7 (Limit compariso test). Let {a } ad {b } be two sequeces of positive umbers. The a) if lim b) if lim c) if lim a 0 b c > 0, a ad b both coverge or diverge together; a b 0 ad b coverges, the a coverges. a Proof. (a) As lim b ad b diverges, the a diverges. Thus, for N, a c > 0, for ɛ c b > 0, there exists N N such that N a c b < c. c a b c c or equivaletly cb a 3cb. Hece the coclusio follows from the compariso test. a b) Give that lim 0. Hece for ɛ b, there exists N N such that or equivaletly, N a b < N a b. Thus the desired coclusio follows from the compariso test. 5

6 c) Here we are give that lim such that or equivaletly, a b. Hece for ay real umber M > 0, there exists N N N a b M N a Mb. Thus if b diverges, the a diverges by compariso test. /// Examples..8. ( + ) ) Cosider the series + ( + ). Here a + ( + ). Let b. The a ( + ) b as. Further, diverges. Thus by limit compariso theorem, the give series diverges. ) Cosider the series. Here a. Let b. The a b coverges ad hece the give series coverges.. Further, 3) Cosider the series e. Here a e ad b. The a b e 0 as. Further, coverges ad hece the give series coverges. 4) Cosider the series e Further,. Here a e coverges ad hece the give series coverges. Theorem..9 (Ratio test). Let a be a series of real umbers. Let The a lim if a) a coverges absolutely if A < ; b) a diverges if a > ; c) the test fails i all other cases. a + a ad b. The a b e 0 as. ad A lim sup Proof. a) If A <, choose B such thata A < B <. The there exists a ɛ > 0 such that B A + ɛ ad also N N such that a + a B for all N. Further, for ay k N, a N+k k a N i a N+i a N+i 6 a + a k B B k. i.

7 Thus a N+k B k a N, k N. But a coverges. a N B k < as B <. Thus by compariso test, the series k0 b) If a >, choose b such that < b < a. There exits N N such that Further, for ay k N, a N+k k a N a N+i k a N+i b b k. Thus a N+k a N, k N. But, as b >, the series a diverges. i i a + a b for all N. a N b k diverges. Thus, agai, by the compariso test, k0 c) Case: a A Cosider the series. Here lim a + a. But diverges. For the series a +, which coverges, agai lim. a Case : A > If we cosider the series the A > ad the series diverges. If we take The it is easy to see that the series coverges as s ( 5 ) + ( 5 )3 + ( 5 ) s + ( 5 ) + ( 5 ) ( 5 ) + ( 5 ) + ( 5 ) But A. Similarly oe ca costruct examples whe a <. /// Examples..0. a) Cosider the series!. Here a + a ( + )+ ( + )!! ( + ) ( + ) e, which is greater tha. So a A e >. Thus the give series diverges. b) Cosider the series x, x R. Here! 0 a + x+ a ( + )!! x x + 0. Therefore a A 0 <. Thus, for all x R, the give series coverges. Theorem.. (Root test). Let a be a series of real umbers. Let A lim sup a. The a) the series coverges absolutely if A < ; 7

8 b) the series diverges if A > ; c) the test fails if A. Proof. a) If A <, choose B such that A < B <. The there exists N N such that a < B for all N. This implies a < B for all N. As B <, the series coverges by compariso test. b) If A >, there exists ifiitely may N such that a >. But this implies that a > for ifiitely may values of ad hece a N 0, i.e., a diverges. c) Cosider the series. Here A ad the series diverges. O the other had, for the series, agai A, but the series coverges. /// Examples... ) Cosider the series x, x R. Here a x. Therefore, x x x. Thus the series coverges for x < ad diverges for x >. ) Cosider the series x, x R. Here a x. The, a x 0. Thus the series coverges for ay x R. 3) Cosider the series a, where a 4 is odd. The lim sup a. Therefore the is eve series coverges. 4) The series 3 ( ). The it is ot difficult to see that lim sup a /3. However ratio test fails i this case. Remark.. We ote that the root test is stroger tha the ratio test. for example, take the series a where The it is easy to see that a { odd + eve lim sup a + a, but lim sup a / /. So the root test implies that the series coverges but ratio test is icoclusive. Alteratig series: Defiitio..3. A alteratig series is a ifiite series whose terms alterate i sig. Theorem..4. Suppose {a } is a sequece of positive umbers such that (a) a a + for all N ad (b) lim a 0, 8

9 the the alteratig series ( ) + a coverges. Proof. Cosider the partial sums with odd idex, s, s 3, s 5,.... Now, for ay N, s + s a + a + s (by (a)). Thus the sequece {s } forms a o-icreasig sequece. Also, otice that s (a i a i ) + a. i Sice each quatity i the parethesis is o-egative ad a > 0, the sequece {s } is bouded below by 0. Hece {s } is coverget. Now, cosider the partial sums with eve idex, s, s 4, s 6,.... For ay N, s + s + a + a + s (by (a)). Thus the sequece {s } forms a o-decreasig sequece. Further, s a (a i a i+ ) a a, which meas that s is bouded above by a. Therefore, {s } is coverget. Let L lim s ad M lim S. By ((b)), i 0 lim a lim (s s ) L M. Thus L M ad hece the alteratig series ( ) + a coverges. /// Examples..5. ) Cosider the series ( ) + /. Here a / as. Hece the above theorem does ot apply. Ayhow, oe ca show that the series diverges. ( ) + ) Cosider the series. The a s of this series satisfies the hypothesis of the above theorem ad hece the series coverges. Examples..6. ) The series ( ) + coverges coditioally. ) The series ( ) coverges coditioally. The followig is a more gereal test tha the previous theorem. 9

10 Theorem..7. (Dirichlet test) Let {a } ad {b } be sequeces of real umbers such that. the sequece s a k is bouded, k. the sequece b is decreasig ad b 0. The the series a b coverges. Proof. Sice A is bouded, there exists M > 0, such that A M for all. Now ote that a b + a b a b s (b b ) + s (b b 3 ) +...s (b b ) + s b Sice b is decreasig, b b + 0. Therefore a b + a b a b Mb Sice b 0, for ay ɛ > 0, we get N such that b ɛ for all N. Now we ca easily see, m a k b k M b Mɛ Therefore, by Cauchy s test, the series a b coverges. /// Examples..8. ) Cosider the series cos π log. Here take a cos π ad b log. The A cos π k (check the first 4 terms ad the use periodicity of cos x) ad b decreases to 0. Hece the series coveges. I this case we ca see that the series does ot coverge absolutely (apply Cauchy s test). ) e!. Take b (log ) ad a (log ) e!. The b decreases to 0. To show the boudedess of the partial sums of a, we ca apply Ratio test to see that the series a coverges. Hece the sequece of partial sums coverge ad so will be bouded. Therefore by Dirichlet test the series a b coverges. Theorem..9. (Itegral Test). If f(x) is decreasig ad o-egative o [, ), The f(x)dx < f() coverges. Details of this theorem will be doe after covergece of improper itegral. 0

11 .3 Problems. If the terms of the coverget series a are positive ad forms a o-icreasig sequece, the prove that lim a 0.. If 0 a ( 0) ad if 0 x, the prove that a x coverges. 3. Test the covergece of the series () (log ) p 4. Determie which of the followig series diverges () (log ) x R. x (a) log 3/ (b) (log ) 3/ (c) + (d) (e) (f). 5. Determie which of the followig series coverges (a) (b)! 0 (c) ( ) (d) (log ) (e)! ( + )! 6. Test the covergece of the ifiite series () 0 ( π ) si () 7. Test the covergece of the ifiite series l( 3 ) (3) ( ) si (4) si( ) ()! ( 3 + ) () si π (c) (!) ( + )[( + )!]

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