Chpter Ifiite Series Pge of Sectio E Rtio Test Chpter : Ifiite Series By the ed of this sectio you will be ble to uderstd the proof of the rtio test test series for covergece by pplyig the rtio test pprecite the limittios of the rtio test E Proof of the Rtio Test The rtio test c be used for testig covergece of series. It depeds o the limitig vlue of the rtio of (+)th term to the th term. Rtio Test (.4). Let be series where is rel d positive. Let + lim L [Rtio of the (+)th term to the th term] (I) If L < the the series ( ) coverges (II) If L > the the series ( ) diverges (III) If L the test fils d we cot coclude whether the series ( ) coverges or diverges. Note: Wht does the Rtio test me i everydy lguge? The rtio test is divided ito three prts. Prt (I) sys tht if the limitig vlue of (+)th term divided by the th term of give series is less th the the series coverges. Wht does prt (II) me? It mes tht if the limitig vlue of (+)th term divided by the th term of give series is greter th the the series diverges. Wht does prt (III) me? If the limitig vlue of (+)th term divided by the th term of give series is equl to the we cot sy whether the series coverges or diverges. The test fils. The proof of prt (I) is difficult becuse it requires kowledge of iequlities, limit of sequece, geometric series, compriso test etc. Also from the outset it is difficult to kow why the geometric series will ply prt i the proof of the rtio test. The oly clue is the iequlities L<, L > d L i the sttemet of the rtio test. Proof of (I). We ssume tht lim + L where L < By the defiitio of limit of sequece (.) there is turl umber such tht ε > 0 (.) lim( x ) For we hve + L < ε for ll L ε > 0 such tht x L < ε
Chpter Ifiite Series Pge of + L < ε Becuse x L < ε mes + < L + ε L ε < x< L+ ε Let r L+ ε where L < d we choose ε > 0 so tht r L+ ε < We c lwys fid such ε becuse L <. Substitutig r L+ ε ito the bove iequlity we hve + < r + < r( ) (*) Wht does this iequlity me? It mes tht the (+)th term is less th r times the th term. Sice this iequlity (*) holds for ll so similrly for +, + 3, + 4,... we hve [ By (*)] < r < r r + + r [ By Above] < r < r r + 3 + 3 r 3 [ By Above] < r < r r + 4 + 3 4 r Addig ll these together we hve 3 4 + + + +... < r + r + r + r +... + + + 3 + 4 3 4 r+ r + r + r +... [Fctorizig] Sice the modulus of the commo rtio Geometric Series ( r ) (**) r < (see ) therefore by (.3) the geometric series r coverges. Hece the bove series r coverges becuse is costt. Sice the series with lrger terms ( r ) coverges therefore by the compriso test (.) the series (o the Left Hd Side of (**)) with smller terms + + + + lso coverges (.3) ( r ) + + + 3 + 4... is coverget if r < (.) If 0 b the ( b ) is coverget ( ) is coverget Wht re we tryig to prove?
Chpter Ifiite Series Pge 3 of We eed to show tht ( ) coverges. How? We hve lredy show tht the series coverges from +, so we c split the ifiite series, ( ) + owrds. We hve, ito the first fiite terms plus the remiig terms from ( ) + + 3 + 4 +... + + + + + + + 3+... Fiite Number of Terms Coverges by Above Does this series coverge? Yes becuse we hve fiite umber of terms plus series which we hve show coverges therefore the whole series, Hece we hve show tht if ( ), coverges. L < the the series coverges. Try the proofs of prts (II) d (III) for yourself. It does tke time to proof somethig for yourself but it is excellet wy of lerig mthemtics. If you do get stuck the exmie the proofs below. Proof of (II). We eed to prove tht if Agi we ssume tht + L > the the series diverges. lim L where L > By the defiitio of limit of sequece (.) there is turl umber such tht ε > 0 + L < ε for ll Becuse x L < ε mes + > L ε L ε < x< L+ ε Sice L > we choose ε > 0 such tht L ε > By substitutig this ito the bove we hve + > + > > 0 for ll > 0 becuse we re cosiderig positive terms. Sice positive (+)th term is greter th the positive th term therefore lim( ) series diverges. Hece we hve show tht if (.) lim ( x ) L > the the series 0 [Not Zero]. By (.6) the diverges. L ε > 0 such tht x L < ε (.6) If lim ( ) 0 the ( ) diverges
Chpter Ifiite Series Pge 4 of Proof of (III). Need to prove tht if L the the test fils d we cot coclude whether the series ( ) coverges or diverges. How do we prove the test fils for L? We c give exmples of series oe of which coverges d the other diverges such tht i both cses + L lim We first give exmple of series which diverges. Cosider. Wht is + equl to? Replcig by + gives + + + + Ivertig the Secod Frctio d Multiplyig + [ Multiplyig] + We hve + + L lim lim Substitutig for + Dividig Numertor d lim Deomitor by + + 0 With we hve the series which diverges becuse it is the well estblished hrmoic series. Now cosider. Wht is + equl to? + + + ( + ) ( ) Ivertig the Secod + Frctio d Multiplyig + x x Becuse ( + ) + y y
Chpter Ifiite Series Pge 5 of + + L lim lim Substitutig + + Dividig Numertor d lim + Deomitor by + 0 The series ( ) coverges becuse this is the p-series with p >. Hece i ech cse L d diverges but coverges. We hve show tht the rtio test fils whe L. To determie whether series coverges or ot it is geerlly esier to pply the rtio test the usig the compriso test becuse you do t eed to fid series to compre with. The limitig vlue of the rtio test L lim + is similr to the limitig vlue of the limit compriso test L lim b but with the compriso test you re comprig with other series b d there re differet coditios o L. However it does deped o the series d with ttemptig plety of exmples i this field you will become fmilir with which test to use for prticulr series. Also the limittio of the rtio test is tht it fils whe L. The rtio test is ofte clled d Alembert rtio test fter the Frech mthemtici Je d Alembert. d Alembert ws bor i Pris i 77 d he studied mthemtics t College des Qutre Ntios which hd excellet mthemtics librry. d Alembert studied my subjects such s lw, medicie, theology, etc he eve qulified s dvocte but it ws his love for mthemtics tht hd priority. Fig Je d Alembert He worked t the Pris Acdemy of Sciece d the Frech Acdemy ll his life. His work o limits led to the rtio test for the covergece of series. The rtio test is med fter him d Alembert s rtio test. Next we pply the rtio test to prticulr exmples. The pplictio of this method is geerlly strightforwrd but c ivolve lrge umber of lgebric steps.
Chpter Ifiite Series Pge 6 of E Applyig the Rtio Test Exmple 5 Determie whether the followig series! coverges or diverges. Solutio. C we pply the rtio test? Yes becuse we oly eed the terms of the series to be positive d sice! > 0 therefore we c use the rtio test. Wht does! me?! 3 4... +! 3 4... + Let the wht is + equl to?! Replcig with + ito gives! + +! ( ) Wht do we eed to fid i order to use the rtio test? The limitig vlue of + Substitutig for +!! We hve ( + ) [ d ]! Ivertig the Secod (! ) + Frctio d Multiplyig!! 3 4... 3 4... + + L + lim + Ccellig All the Commo Fctors + lim 0 Substitutig + + Sice L 0< therefore by the rtio test (.4) (I) If L < the the series ( ) coverges the give series! coverges. Remember the rtio test sys tht if L < the the series uder cosidertio coverges. We oly eed to check the vlue of L. The met is i evlutig L.
Chpter Ifiite Series Pge 7 of Exmple 6 Discuss the covergece or divergece of 0 (! ) Solutio. How c we test the give series for covergece? Sice the th term does ot coverge to 0, tht is we c coclude by (.6) tht 0 (! ) ( ) lim! 0 rtio test s follows: Let! the ( )! Substitutig these ito + ( ) +! +! diverges. We c cofirm this by pplyig the + ( ) + gives 3 4... + Ccellig + 3 4... Commo Fctors Hece + L lim lim ( + ) + Sice L > therefore by the rtio test (.4) (II) If L > the the series ( ) diverges the give series 0 (! ) diverges. The ext exmple is ot difficult but just ivolves lot of lgebr. Of course if you do ot kow the result of the hit, lim + e, the the questio of testig the give series for covergece is difficult. Exmple 7 Show tht the followig series! coverges. [Hit: lim + e ]. Solutio. C we pply the rtio test?! Yes becuse is positive for ll. Let!. Wht is + equl to? Replcig with + gives (.6) If ( ) lim 0 the ( ) diverges
Chpter Ifiite Series Pge 8 of ( + ) ( + )! + + Let + L lim (*) Let s ivestigte the terms iside the brckets. ( +! ) +! Substitutig for + ( ) + (! ) + ( ) ( +! ) + ( + )! ( + ) ( + ) ( ) [ d ] + + Ivertig the Secod! Frctio d Multiplyig + ( + )! Becuse + +! Ccellig ( + )'s + + x y Remember y x + + Becuse + + + Rewritig x x + + Substitutig ito (*) we hve + L lim + < [By Hit: lim e] + e lim + Sice L < therefore by the rtio test (.4) (I) If L < the the series ( ) coverges! the give series coverges.
Chpter Ifiite Series Pge 9 of Remember for usig the rtio test we eed to evlute L. Oce we hve prticulr vlue of L the we c decide o the questio of covergece of give series. Exmple 8 Determie whether the followig series coverges + Solutio Let +. Wht is Replcig with + i + equl to? + gives ( + ) + + + + + Substitutig these ito + gives + + + + We hve L + + + + + 3 + + + 3 + lim + [ Simplifyig] [ Expdig Brckets] + 3 + lim Substitutig for + 3+ + 3 Dividig Numertor lim + 3 + d Deomitor by L + Does the give series coverge or diverge? By pplyig the rtio test with L + (.4) (III) If L lim the test fils we cot swer the questio whether the series coverges or diverges. How c we test the give series? Well we c fid the limit of the th term of the give series
Chpter Ifiite Series Pge 0 of + + Dividig Numertor lim lim d Deomitor by Sice the th term coverges to 0[Not Zero] therefore by (.6) the give series + diverges. For Exmple 8 bove we should hve oticed from the outset tht the th term does ot coverge to zero therefore the series diverges. Remember the rtio test does ot work whe L. We hve to pply some other test i this cse. Exmple 9 x be rel sequece. Show tht if Let the ( x ) coverges. Solutio C we pply the rtio test? 0 < x+ < x Yes becuse x is positive for ll the turl umbers. Let x the + ( x ). We hve + + + ( x+ ) ( x ) + ( x ) ( x ) + + x [ Usig the Rules of Idices] x + x+ ( ) x Sice 0 < x+ < x therefore x + < x x+ + Let r < d substitutig this ito ( ) we hve rx x L lim + lim ( rx ) Substitutig rx How do we evlute lim rx +? r < ( r ) Sice therefore lim 0. (.6) If ( ) lim 0 the ( ) diverges + + + +
Chpter Ifiite Series Pge of Also becuse the sequece ( x ) is mootoic (decresig) sequece d is bouded by x therefore by the Mootoic Sequece Theorem (.??) ( x ) coverges. Let M lim ( x ) the from bove we hve ( ) ( ) Sice L 0< by the rtio test (.4) (I) If the give series x L lim r x+ lim r lim x+ 0 M 0 0 becuse r< L < the the series coverges coverges provided tht 0 < x+ < x for ll. There re lots of exmples i Exercise e for you to tckle. Try to ttempt ll questios i this exercise. The oly wy you will become fmilir with the rtio test is to do lrge umber of questios eve if your lecturer does ot ssig y. This is the oly wy of lerig mthemtics. SUMMARY Rtio Test (.4). Let be series where is rel d positive. Let + lim L [Rtio of the (+)th term to the th term] (I) If L < the the series ( ) coverges (II) If L > the the series ( ) diverges (III) If L the test fils d we cot coclude whether the series ( ) coverges or diverges. (.??) A bouded mootoic sequece coverges