MATHEMATICS I COMMON TO ALL BRANCHES

Transcription

1 MATHEMATCS COMMON TO ALL BRANCHES

2 UNT Seqeces ad Series. Defiitios,. Geeral Proerties of Series,. Comariso Test,.4 tegral Test,.5 D Alembert s Ratio Test,.6 Raabe s Test,.7 Logarithmic Test,.8 Cachy s Root Test,.9 Alteratig Series,.0 Series of Positive ad Negative Terms,. Power Series,. Uiform Covergece ad Weirstrass s M-test.. DENTONS. Seqece. A fctio whose domai is the set of atral mbers ad rage a sbset of real mbers is called a real seqece or simly a seqece. Symbolically, if : Natral Nmbers real mbers, the is a seqece. t is deoted as { }, where is the -th term of the seqece or simly,,... Ths,,, 4... is a ifiite seqece.. Mootoic seqece. f i the seqece { }, (i) for all or, (ii) for all, the the seqece is said to be mootoic icreasig (for i) or mootoic decreasig (for ii) seqece. Ths, if, { } is mootoic icreasig seqece ad if, { } is mootoic decreasig seqece.. Boded seqece. f corresodig to the seqece { } there eists a fiite mber M sch that M for all the seqece is said to be boded above ad if M for all, the seqece is said to be boded below. f a seqece is boded both above ad below it is said to be boded. Ths, RST UVW is boded above ad the er bod is ad RST U V W is boded below ad the lower bod is 0. V. Coverget seqece. A seqece { } is called coverget ad is said to have a limit l, if R S U V Lt l. Ths, is coverget ad the limit is. T W V. Diverget seqece. The seqece { } is called diverget if C-4\N-MATH\jc-.m5

3 4 Tetbook of Egieerig Mathematics Lt or. Ths { } is diverget. V. Oscillatory seqece. f a seqece either coverges or diverges to or, the the seqece is said to be oscillatory. Moreover, if it is boded, it is said to be oscillate fiitely or if it is boded it is said to be oscillate ifiitely. Ths, {,,,,,,...} oscillates fiitely ad {( ). } oscillates ifiitely. Now we state the ecessary ad sfficiet coditios that a seqece { } may be coverget as follows : (i) Either, the seqece is mootoic ad boded i.e., mootoic icreasig ad boded above or mootoic decreasig ad boded below. (ii) Or else, the seqece satisfies Cachy s coditio amely, Give ad re-assiged small ositive qatity, we ca determie a ositive iteger N, sch that <, wheever N, beig ay ositive iteger. R Eamle. Prove that the seqece S U T V is coverget. W Let, ( ) Now < 0 for all. [( ) ] ( ) The < for all i.e., { } is mootoic decreasig. Also, > 0 for all i.e., { } is boded below by 0 Hece the give seqece is coverget Also, Lt Lt Lt / 0. / Eamle. Show that the seqece {( /) } is mootoic icreasig. Let Assme that > H K if, > H K if, / > /( ) C-4\N-MATH\JNC-.m5

4 if, > if > / if ( / ) > / which is tre by Berolli s ieqality. [or every ositive iteger ad > ( 0) ( ) > ]. Hece the give seqece is mootoic icreasig. Eamle. Prove that the seqece R S T 4 4, 4( ) 4 7 ( ) 5 Now, > 0 ( )( ) U V W Seqeces ad Series 5 is boded ad mootoic icreasig. > or, < Pttig,,,... we get < < <... < <... So { } is mootoic icreasig. Now, < 4 ad 7 7 < < 4. Hece { } is boded. Eamle 4. Discss the covergece of the seqece { } where... Now, > 0 for all. ( )( ) for all i.e., < i.e., < <... < <... So { } is mootoic icreasig. Also becase each of,..., is less tha /, it follows that <. for all. { } is boded above. Hece, { } is mootoic icreasig ad boded above ad is, therefore, coverget. V. fiite series. f { } be a give real valed seqece, the a eressio of the form is called a ifiite series. symbols it is geerally writte as or. C-4\N-MATH\JNC-.m5

5 6 Tetbook of Egieerig Mathematics Whereas m is called a fiite series assmig that the terms,,... are all zero. V. Seqece of artial sms of a ifiite series. Let s associate to the ifiite series a seqece {s } defied by s... The seqece {s } is called the seqece of artial sms of the give series. X. Covergece of a ifiite series. A ifiite series Σ is said to coverge, diverge or oscillate accordig as its seqece of artial sms {s } coverges, diverges or oscillates. case {s } coverges to s, tha s is called the sm of the series Σ ad we shall write s Σ. Mathematically, Σ coverges if Lt fiite Σ diverges if Lt s s or Σ oscillates if {s } oscillates (fiitely/ifiitely). i.e., {s } does ot have a iqe limit. X. Series of ositive terms. A ifiite series i which all the terms are ositive after some terms is called ositive term series e.g.,... is a ositive term series.. GENERAL PROPERTES O SERES P : Covergece of a series remais chaged by the relacemet, iclsio or omissio of a fiite mber of terms. P : A series remais coverget, diverget or oscillatory whe each term of it is mltilied by a fied mber other tha zero. P : A series of ositive terms either coverges or diverges to i.e., omittig the egative terms, the sm of first terms teds to either a fiite limit or. P4 : Every fiite series is a coverget series. Eamle. Show that if a series Σ is coverget the Lt 0. Let s deote the th artial sm of the series Σ. The Σ is coverget {s } is coverget. Lt s (fiite) Also, s Lt s s Bt s s Lt Lt ( s s ) Lt s Lt s s s 0 The coverse of the above reslt (Eamle ) is ot always tre i.e., Lt 0, imlies that Σ may or may ot be coverget. or eamle, diverges, thogh Lt Lt 0. C-4\N-MATH\JNC-.m5

6 Eamle. The geometric series... (i) Coverges if < < (ii) Diverges if (iii) Oscillates fiitely if (iv) Oscillates ifiitely if < (i) < < i.e. < Let s... ( ) Lt s {s } is coverget, hece the give series coverges (ii) i.e., ad > or, s... to terms Lt s {s } is diverget hece the give series is also diverget or > s... Seqeces ad Series 7 H Q Lt 0as < K Lt s Q Lt, as > H K {s } is diverget hece the give series is also diverget. (iii) s... to terms. or 0 if is odd or eve Lt or 0. s {s } oscillates fiitely ad hece the give series oscillates fiitely. (iv) < Here, < > Let, the > Now s... ( ) or if eve or odd. C-4\N-MATH\JNC-.m5

7 8 Tetbook of Egieerig Mathematics Lt s or (Q as ) {s } oscillates ifiitely ad hece the give series oscillates ifiitely.. COMPARSON TEST. f two ositive term series Σ ad Σv be sch that (i) Σv coverges ad (ii) v, the Σ also coverges. (Note. f the relatio v holds for > m, the the first m terms of both the series ca be igored which will ot affect their covergece or divergece).. f two ositive term series Σ ad Σv be sch that (i) Σv diverges ad (ii) v, the Σ also diverges.. f two ositive term series Σ ad Σv be sch that Lt l, l beig a o-zero fiite v qatity the Σ ad Σv either both coverget or both diverget. articlar, if l 0 ad Σv coverges, the Σ also coverges, if l ad Σv diverges, the Σ also diverges. Amog,,, is widely sed. V. A imortat series for comariso is the Harmoic series of order i.e.... coverges if > ad diverges if. Eamle. Test the covergece of the series T of,, 5,,,... / ( )( ) ( / )( / ) Let v. Let s comare with Σ ad Σv / Lt Lt 0 ad fiite v ( / )( / ). Q Σv is of the form with > Σv is coverget Σ is coverget. Eamle. Test the covergece of the followig series : e j e je j C-4\N-MATH\JNC-.m5

8 Seqeces ad Series 9 e / j Let v. Lt Lt v e / j 0 ad fiite bt Σv diverges H, for diverget K Σ also diverges. Eamle. Eamie the covergece of the followig series : Here Let v Lt v ad fiite bt Σv is coverget Σ is also coverget d id i d i d i d / i / d / Lt d / i 0 H, > i / for coverget K PROBLEMS Test the Covergece or Divergece of the followig series : H 4 6. K ( ) 8. L O NM QP C-4\N-MATH\JNC-.m5

9 0 Tetbook of Egieerig Mathematics ANSWERS. Diverget. Diverget. Coverget 4. Coverget 5. Coverget 6. Coverget for > ad diverget for 7. Coverget for > /, diverget for /. 8. Coverget..4 NTEGRAL TEST A ositive term series Σf() where f() decreases as icreases the the series Σ ad the itegral z f d coverge or diverge together i.e., for coverget the vale of the itegral mst be fiite ad for diverget the vale of the itegral mst be ifiite. Proof. Give that f() is a mootoic decreasig fctio. Let r be a ositive mber sch that r r. f(r ) f() f(r) y r f() r tegratig i [r, r ] z z z r r r r r r ( ) r zr r r f d r d f d d ( ) r...(i) This ieqality is tre for the iterval [r, r ]. Now cosider the itervals [, ], [, ],... [r, r ],..., [, ]. The totality of all the itervals leads to the followig ieqality. z z... f ( ) d f ( ) d... f( ) d.... i.e., z s f( ) d s (s Σ ) Takig the limit z Lt s s...(ii) z z (a) f ( ) d ad Lt s will be fiite mbers. So {s } is a coverget seqece ad Σ is coverget. z z (b) f ad Lt s will be. So {s } is a diverget seqece ad Σ is diverget. 0 r r z f() Hece Σ ad z f d ( ) coverge o diverge together. Note. Σ ad z ( ) k f d, k also coverge or diverge together. C-4\N-MATH\JNC-.m5

10 Seqeces ad Series Eamle. Show that the harmoic series of order.... Coverges for > ad diverges for 0 <. Here f(), for, f() is ositive, ad mootoic decreasig. tegral test is alicable. z The above series will coverge or diverge accordig to d is fiite or ifiite. Case (i) Case (ii) z z d k k z z d k k d k Lt Lt k (fiite), for >, for 0 < < d Lt Lt l k k Σ coverges if > ad diverges if 0 <. Eamle. Test the covergece of the followig series. 4 Here, f(). 4 or, f() is ositive ad mootoic decreasig. tegral test is alicable. z z k Now, f( ) d d Lt 4 4 d k z 4 k Lt l ( ) Lt k k Σ is diverget. 4 [l ( k ) l 4] PROBLEMS. Test the Covergece of the followig series :. l C-4\N-MATH\JNC-.m5

11 Tetbook of Egieerig Mathematics (l ) 4. l l (l ) l (l (l )) ANSWERS. Div. 5. Cov. for >, div. for. 8. Cov.. Div. 6. Div. 9. Div.. Cov. for >, div. for 0. Cov. 4. Div. 7. Cov..5 D ALEMBERT S RATO TEST f Σ be a ositive term series ad Lt m the (i) Σ is coverget if m <. (ii) Σ is diverget if m >. (iii) No coclsio ca be draw if m i.e., the test fails ad the other method is reqired. Proof. (i) Let m > ad fiite for m < m, there eists a atral mber l sch that Takig v, we get l m > > for all l. v m > for all l v Sice Σv is coverget, therefore, by comariso of ratios test, Σ is coverget. (ii) Let m < ad there eists a atral mber l sch that < for all l l < for all l <. C-4\N-MATH\JNC-.m5

12 Seqeces ad Series Lt 0 Σ is diverget. or m, we ote that both the series have Lt, bt the first series is coverget ad the secod is diverget. So the series Σ may coverge of diverge. Hece the test fails for m. Eamle. Discss the coverget of the followig series : (i) (ii)... (i) Here ad ( ) [ ] Lt Lt [( ) ] Lt [ / ] [( / ) / ] [ / ] Lt ( / ) / > Hece by ratio test, Σ is diverget. (ii) Here ad ( )( ) Lt Lt Lt ( ) ( ) H K ( ) Lt. H K ( ) / < Σ is coverget by ratio test. Eqivalet Statemet f Σ be ositive term series ad Lt m the (i) Σ is coverget is m > (ii) Σ is diverget is m < (iii) No coclsio ca be draw if m. Eamle. Test for covergece for C-4\N-MATH\JNC-.m5

13 4 Tetbook of Egieerig Mathematics Here, ( ) Lt Lt Lt ad / /. / / / ( ). Σ coverges accordig to < ad diverges accordig to >. f, Let v By comariso test, / / ( ) ( / ) Lt Lt 0 ad fiite. v / Q Σv is coverget, Σ is also coverget. Hece, by ratio test the give series is coverget for ad is diverget for >. Eamle. Test the covergece of the followig series : (i), (ii) (i) Here ( )! 4, ( ) ( ) ( ) ( ) 4 [( ) ] Lt Lt. 4 ( ) Lt.( / ) Lt ( / ) Hece by the ratio test Σ coverges. (ii) Here Lt. L NM ( )! ( )! Lt. ( / ) [ / ] ( / ). / 4 ( / ) By D Alembert s ratio test, Σ coverges. O QP. < ( )! ( )!. Lt ( )( ). 0 < C-4\N-MATH\JNC-.m5

14 Eamle 4. Eamie the covergece of the followig series :! 5! 5! ( )! Here ( )! ( ) ( )! Lt Lt ( )!( ) Lt ( )( ) ( ). ( / ). Lt Lt ( / ) By D Alembert s, ratio test the give series diverges. Seqeces ad Series 5 PROBLEMS Discss the covergece of the followig series : ! ( > 0)!! 4! 7. α α α α α α β ( )( ) ( )( )( )... ( β )( β ) ( β )( β )( β ) ! e e e... e ANSWERS. Div.. Cov.. Cov. 4. Cov. for < ad div. for 5. Cov. 6. Cov. 7. Cov. if β > α > 0, div. if α β > 0 8. Cov. 9. Cov. 0. Cov.. Div.. Div. C-4\N-MATH\JNC-.m5

15 6 Tetbook of Egieerig Mathematics.6 RAABE S TEST The series Σ of ositive terms is coverget or diverget accordig as Lt R S T U V W > or < The test fails if the limit. Geerally, this test is alied whe D Alembert s ratio test fails ad also whe i the ratio test does ot ivolve the mber e. Proof. Let Lt R S T (i) m >, for m U V W m be fiite < m, there eists a atral mber l sch that m > > ( ) > m for all l, m Ths [ ( ) ] > l i.e., l l > m ( l... ) for all l, or m l l... m >... for all l. This shows that the seqece of artial sms {s } is boded ad so, beig mootoic it coverges. Therefore Σ coverges for fiite m >. (ii) f m <, there eists a atral mber l sch that < i.e., < ( ) for all l i.e., < ( ) l l i.e., l l < for all l i.e., (l l ). < for all l Sice is diverget ad l l reresets ositive real mber, Σ is diverget. or m, both the series l C-4\N-MATH\JNC-.m5

16 Seqeces ad Series 7 ad (l ) have Lt, bt the first series is coverget whereas the secod series is diverget. Hece for m, the test fails. Note. (De Morga s Test or Bertrad s Test) Let Σ be a series of ositive terms sch that Lt NM log l. The (i) Σ T QP W coverges if l >, (ii) Σ diverges if l <. [This test may be alied whe both D Alembert s ratio test ad Raabe s test fails]. Eamle. Discss the covergece of the series : ( > 0) Neglectig the first term, we have ( ). ad ( ) ( ) ( ) / Lt Lt. Lt. / By D Alembert s ratio test, the series coverges for < ad diverges for >. The test fails for. We aly Raabe s test for. Now, RL S M Lt Lt Lt / < By Raabe s test the series diverges. Hece the give series coverges for < ad diverges for. Eamle. Test the covergece of the series ( ) ( ). Here, ( ) ( ) ( ).... O P U V d-4\n-math\jnc-.m5

17 8 Tetbook of Egieerig Mathematics Now, Lt by D Alembert s, ratio test, the series is coverget if <, i.e., < ad diverget, if >, i.e., >. f, the test fails. Let s aly Raabe s test. Here, R S T R S T U V U V W ( 6 ) 6 / ( ) ( / ) Lt W 6 4 i.e., > The series is coverget i this case. f, all the terms beig reversed i sig as comared to the case whe, the series remais coverget. Ths the give series is coverget whe, ad diverget whe >. Note. The series is the easio of si ().. Test the covergece of the followig series : PROBLEMS ( ) ( ) ( > 0) ANSWERS. Div.. Cov.. Cov. 4. Cov. for < ad Div. for. 5. Cov. for <, ad Div, for. 6. Cov. if, Div. if >. 7. Cov. if <, Div. if. 8. Div..7 LOGARTHMC TEST A ositive term series Σ coverges or diverges accordig to L M Lt log > or < N Q bt the test fails for the limitig vale. O P d-4\n-math\jnc-.m5

18 Seqeces ad Series 9 Note. The test is alied after the failre of D Alembert ratio test ad geerally whe This test is a sbstitte for Raabe s test. ts roof is similar to that of Raabe s test. Eamle. Test the covergece of the series i.e., > e.!!... 4! Here, ( )!, ( ) ( ) ( ) Lt Lt. ( ) Lt ( / ).( / ) ( / ).( / ). e e. Lt e L N M H ivolves e. a e K By D Alembert s ratio test, the series coverges if e < i.e., < e ad diverges if e > a O Q P f e the test fails. Now whe e, ( / ) ( / ) log ( ) log ( /) ( ) log ( /) log e ( ) [log ( /) log ( /)] ( ) L 4 8. H. NM... K H K O.... Q P ( ) L... O... NM QP H K... Lt Lt log H... K Lt... / < H K So by Logarithmic test, the series diverges. Hece, the give series coverges if < e ad diverges if e. /e. d-4\n-math\jnc-.m5

19 0 Tetbook of Egieerig Mathematics Eamle. Test the covergece of the series Here, ( )! ( )!, ( ) ( )! Lt Lt. e By D Alembert s ratio test the give series coverges or diverges accordig as < /e or > /e. or /e the test fails. log.. e log ( )log N L NM L NM RST R S T L M. ( ) log H K log H O K UVW QP H O Q P ( ) Lt log <. Hece by the logarithmic test the series diverges for /e. Ths the give series coverges or diverges accordig as < /e or /e. UO KV W QP Test the covergece of the followig series : PROBLEMS !! 4! 4 e!. e!. e 4!. e e e 4 e e !! 4! 5!!! 4! 5.!! 4! ANSWERS. Cov.. Div.. Cov. if e, Div. if > /e 4. Cov. if < /e, Div. if /e, 5. Cov. if < e, Div. if e. d-4\n-math\jnc-.m5

20 .8 CAUCHY S ROOT TEST f Σ is a ositive term series ad Lt Seqeces ad Series ( ) / m, the the series is (i) coverget if m < (ii) diverget if m > ad o coclsio ca be draw abot the covergece or divergece if m. Proof. Let m 0. or m < m <, there eists a atral mber l sch that 0 < / < m m for all l. Ths < for all l. H K m Sice H is coverget for 0 < K m < By comariso test is coverget. or m >, there eists, a atral mber l sch that / > > for all l. Lt 0 Σ diverges. f /, there eists a atral mber l. Sch that / > l i.e., > l for all l Lt 0 Σ diverges / or m, we ote that, ad have Lt bt the first series is coverget ad the secod series is diverget. the test fails i this case. Eamle. Test the covergece of the followig series H Here. K / Lt Lt H K < By Cachy s root test, the series is coverget. Eamle. Eamie the covergece of the series [( ) ] [( ) ] / ( ) ( ) Lt Lt Lt / /..... d-4\n-math\jnc-.m5

21 Tetbook of Egieerig Mathematics Lt.. H K / By Cachy s root test, the series coverges for < ad diverges for >. PROBLEMS.. 5. Test the covergece of the followig series. H K H 5 7K H K H K (log ) ANSWERS. Cov.. Cov.. Cov. 4. Cov. 5. Cov. for <, div. for..9 ALTERNATNG SERES : (LEBNTZ S TEST) The alteratig series ( ). Coverges if (i) > for all ad (ii) Lt 0. Here both the coditios (i) ad (ii) mst be satisfied for covergece. f Lt 0, the give series is oscillatory. Proof. f S deotes the artial sm of the give alteratig series the, S 4... [( ) ( 4 5 )... ( ) ] < (Q > for all ) This imlies that {S } is boded above Also, S S S S S S > 0 (Q > > for all ) S > S This imlies that {S } is mootoically icreasig. Hece {S } is coverget i.e., Lt S fiite d-4\n-math\jnc-.m5

22 Seqeces ad Series Also, Lt S Lt ( S ) Lt S 0 Q Lt 0 give Ths Lt S Lt S fiite fiite whe is odd or eve. Hece the give series is coverget. Whe Lt 0, Lt S Lt S, The give series is oscillatory. Eamle. Discss the covergece of the followig series. t is a alteratig series. (i) Sice, (ii) Lt 0 ad ( ) > for all, ( ) > for all H Both the coditios of Leibitz s test are satisfied. Hece the give series is coverget. Eamle. Test the covergece of the followig series t is a alteratig series (i) 5, 5( ) Q 5 > for all. 5( ) > for all. (ii) Lt Lt Lt 5 5 / 5 0 Secod coditio of Leibitz s test is ot satisfied. Hece the give series is Oscillatory. K d-4\n-math\jnc-.m5

23 4 Tetbook of Egieerig Mathematics. Show that the series PROBLEMS H K H 4 5K... is coverget bt the series obtaied from the first by omittig brackets is ot coverget.. Show that the series is coditioally coverget. [Hits. See the et sectio]. Eamie the covergece of the followig series : (i) (iii) (ii) ( ) L N M O Q P (iv) ( ) si ( / ) log log log 4 log 5. ANSWERS. (i) Cov. (ii) Oscillatory (iii) Cov. (iv) Cov..0 SERES O POSTVE AND NEGATVE TERMS The series of Positive terms ad alteratig series are the secial tyes of these series with arbitrary sigs. The series Σ of arbitrary terms is said to be absoltely coverget if Σ is a coverget series. Whereas, the series Σ is said to be coditioally coverget if it is coverget bt does ot coverge absoltely i.e., Σ is diverget. To test Σ is coverget or ot, sitable tests as discssed i the earlier sectios has to be alied. Eamle. Test whether the series is absoltely coverget or coditioally coverget?... 4 This is a alteratig series (i) ad obviosly, ( ) > for all. d-4\n-math\jnc-.m5

24 Seqeces ad Series 5 (ii) Lt 0. Both the coditios of the Leibitz s test are satisfied, so the give series is coverget. Agai, cosider the series with absolte terms Σ... 4 Which is a coverget series, ( > i the Harmoic Series). Ths the give series coverges absoltely. Eamle. Prove that the followig series coverges absoltely ! 5! 7! Cosider the series by takig all the terms ositive 5 7 i.e.,......(i)! 5! 7! Here, ( )!, ( )! Lt Lt 0 for all. ( ) Hece the series (i) is coverget for all. Therefore the give series is absoltely coverget for all. Reslt. Every absoltely coverget series is coverget bt the coverse is ot tre. Let Σ be a absoltely coverget series. Obviosly, i.e., Σ Σ where Σ is coverget. Hece Σ is also coverget. PROBLEMS. Tet the covergece of the series f S is the sm of the coditioally coverget series ( ), the show that the rearraged series.... has the sm S.. Test whether the followig series are absoltely coverget or coditioally coverget? ( ) (i) (ii) ( ) (iii) (iv) d-4\n-math\jnc-.m5

25 6 Tetbook of Egieerig Mathematics ( ) (v) log 4. Prove that the series si si si... Coverges absoltely. 5. Show that the series ( ). (vi)... ( ) log log log 4 log 5 log... Coverges coditioally. ANSWERS. Div.. (i) Coverges coditioally, (ii) Absolte cov. (iii) Coditioally cov. (iv) Coditioally cov. (v) Coditioally cov. (vi) Absolte cov.. POWER SERES A series of the form a 0 a a... a... where a 0, a,... a,... are all costats is called a ower series i which may coverge for some or all vales of. These vales form the iterval of covergece. Also every ower series is coverget for 0. Eamle. Show that the eoetial series Here is coverget for all vales of.!!!, ( )! Lt Lt 0 Hece by D Alembert s ratio test the series coverges for all vales of. Eamle. Show that the logarithmic series Here, ( )., ( ).... a f... is coverget for <. Lt Lt Lt / The series coverges for < ad diverget for >, whe, the series 4 d-4\n-math\jnc-.m5

26 Seqeces ad Series 7 Whe, the series... is diverget (comarig with Harmoic series) 4 Hece the series coverges for <. Eamle. or what vales of the followig series is coverget?!!! Here! ( )!, ( ) Lt Lt Lt ( )! ( ) ( / ).! Lt. e or covergece by D Alembert s ratio test e K J. < < e i.e., e < < e. PROBLEMS or what vales of are the followig series coverget : ( ) ( )... ( r )... r...! r! si si 4 si 9... si ( ) ( ) !.!!...!...! 4! 6! ANSWERS. <. < <. < 4. < 5. < < < 6 8. < < < <. d-4\n-math\jnc-.m5

27 8 Tetbook of Egieerig Mathematics. UNORM CONVERGENCE AND WEERSTRASS S M-TEST ad Let ( ) be a ifiite series of fctios each of which is defied i the iterval (a, b). Let s () () ()... () sm of first terms s() Lt s ( ) The series Σ () is said to be iformly coverget i the iterval (a, b), if for a give > 0, there eists a mber N which is ideedet of, sch that for every i the iterval (a, b), s() s () < for all > N By defiitio, iform covergece imlies covergece for each oit i (a, b) bt the coverse may or may ot hold. Eamle. The series ( )( ) ( )( )... Coverges iformly o [a, ) where a > 0. Here s ()... ( )( ) (( ) ) ( ) ad so, R () Let 0 < <. The R () < for > H. Ths if we take N K H a, we get K r () < for all N ad a > 0. Hece the same N serves for the covergece of the series for all is [a, ) ad so the give series coverges iformly o [a, ), a > 0. Weierstrass s M-Test for Uiform Covergece A series Σ () is iformly coverget i a iteral (a, b), if there eists a coverget series ΣM of o-egative terms sch that () M for all vales of i (a, b). Proof. f ΣM is coverget, for > 0, there eists, N sch that M M... M < for all N ad. or all i (a, b) ( ) ( )... ( ) ( ) ( )... ( ) M M... M M M... M < for N ad Hece, Σ () coverges iformly o (a, b). d-4\n-math\jnc-.m5

28 Also from above, () ()... () < for all N ad o (a, b). Therefore, Σ () also coverges absoltely o (a, b). Eamle. Show that the series Coverges iformly o [a, ], 0 < a < bt ot o [0, ]. O [a, ] where 0 < a <, ( ) a ad ow, to test the covergece of a Let f f a, ( ) a f a / a Lt Lt Lt f ( ) a / ( / ) a D Alembert s ratio test fails Now, we shall aly Raabe s test f ( ) a a a. f a a Seqeces ad Series 9 a a f. ( a a / ) Lt Lt a f (/ a ) > a Σf coverges Therefore by Weierstrass s M-Test, the give series coverges iformly o [a, ] O the otherhad, let the give series iformly coverget o [0, ], the for > 8 eists m, sch that... < m ( m ) ( m ) 8 m ( m) < 8 0 there (by takig m) O ttig it gives <, a cotradictio, therefore the give series is ot iformly m 5 8 coverget o [0, ]. d-4\n-math\jnc-.m5

29 0 Tetbook of Egieerig Mathematics Eamle. Show that the series Σ si coverges absoltely ad iformly o (a, ) 4. where a > 0. or ay > 0 i (a, ), there eists a N sch that 4 0 for all N. Hece the series after a fiite mber of terms cosists of ositive terms. si 4. Sice Lt Lt. 4 < si 4. The series coverges absoltely o (a, ) if a > 0. Also, for M, si Ths, < <. N N si < 4 4 H 4 K for all N Sice Σ4 N H 4K is coverget, therefore, by Weierstrass s M-Test, Σ si /4 coverges iformly o (a, ) where a > 0. PROBLEMS. Show that has a maimm at ad hece or otherwise show that ( ) coverges iformly i (0, ).. Test for iform covergece of the series si (i) o [0, k] for ay k > 0 (ii) si si si cos (iii) ad si (iv) ( ) (v) (vi)! 5! 7! ! 4! 6! ANSWERS. (i) Uiform cov. o [0, k], k > 0 (ii) (vi) Uiform cov. for ay real. ( ) REVEW QUESTONS. Give the differece betwee seqece ad series.. Give a eamle of a mootoically icreasig seqece which is coverget. d-4\n-math\jnc-.m5

30 Seqeces ad Series. What is meat by Oscillatory seqece. 4. State the roerties of series. 5. What is meat by coverget of a series. 6. Uder what coditio, Logarithmic test is alied istead of Raabe s test. 7. Defie iform covergece of a series. 8. State Weierstrass M-Test for Uiform Covergece of a series. 9. Defie absolte ad coditioal covergece with eamles. OBJECTVE QUESTONS. or the seqece.6,.66,.666,... H... the least er bod is 0 K (a) 0 (b) (c) / (d) /. or the seqece,,,,... ( ),..., the greatest lower bod is (a) 0 (b) (c) (d). The limit oit of the seqece 0, ,,,,,... is (a) 0 (b) (c) (d) Does ot eist 4. or the seqece {( ) }, the limit oits are (a) 0, (b) 0, (c), (d) Does ot eist 5. or, the geometric series.... (a) Coverges (b) Diverges (c) Oscillates fiitely (d) Oscillates ifiitely 6. f m is a give ositive iteger, the cosider two series, S... m... ad S m m... (a) S Coverges, S Diverges (b) S Coverges, S Coverges (c) S Diverges, S Coverges (d) S Oscillates, S Diverges 7. The series coverges for (a) (b) (c) > (d) 8. Whe D Alembert s ratio test fails ad ivolves e the (a) Raabe s test is alied (b) Cachy s Root test is alied (c) Logarithmic test is alied (d) tegral test is alied 9. O Alteratig series Σ( ), (a) Cachy s Root test is alied (b) D Alembert s ratio test is alied (c) Leibitz s test is alied (d) tegral test is alied 0. the iterval [ /, /], the geometric series... (a) Coverges (b) Coditioally coverges (c) Absoltely coverges (d) Uiformly coverges d-4\n-math\jnc-.m5

31 Tetbook of Egieerig Mathematics. f Lt a l (fiite), the the iterval of covergece for the ower series a 0 a... a... is a (a) 0 < < l (b) 0 < < /l (c) < < l l (d) l < < l. Every ower series of the form a 0 a... a... is coverget for (a) 0 (b) (c) < (d). The series... coverges absoltely if (a) < 0 (b) > (c) 0 < < (d) < < 4. The series... is 4 (a) Uiformly coverget (b) Absoltely coverget (c) Coditioally coverget (d) Diverget 5. The series is coverget if (a) < 0 (b) 0 < < (c) (d). ANSWERS. (c). (c). (b) 4. (c) 5. (c) 6. (b) 7. (c) 8. (c) 9. (c) 0. (d). (c). (a). (c) 4. (c) 5. (c). d-4\n-math\jnc-.m5