Sequences and Series @ 2.9 ‘The given sequence is 1, 2, Ae “1 the sequence is a bounded below by 0 [and above by 1}. Hence, the sequence is convergent EXERCISE 2.1 _ Show that the sequence {a} converges to 1. 1 x v 1 . Show that the sequence {ahs converges to z 2 . Show that the sequence { 3nt +n +" | converges to 3. e ve 4, Test the convergence of the sequence {e=2l. n+2. 5, Test the convergence of the sequence 2 aah 2n+l 6. Test the convergence of the sequence eco F ANSWERS TO EXERCISE 2.1 4. Converges 5. Converges 6. Converges 2.2 SERIES Definiton2.8 If {u,} bea sequence of real numbers, then the expression iy +, +a Hitt is called an infinite series and it is denoted by Su, or De u, is called the n* term of the series. 2.2.1 Convergent Series Definition2.9 Let u, +i, +, tet, te bean infinite series. Ifs,=u, tu, +--+u,,then s, is called the n® partial sum of the series. Ifthe sequence of partial sums 1. converges to / and it is written as Su ct {s,} converges to /, then we say that the series si, Then is called the sum of the series. Scanned with CamScanner2.10 Engineering Mathematics 2.2.2 Divergent Series Definition 2.10 Tf the se quence of partial sums (s,} of the infinite series diverges, then thes diverges. es X That is, if tims, = 00, then Su, diverges to » and if tims, = ee, then diverges to, nel at y: 2.2.3 Oscillatory Series Definition 2.11 Ifthe sequence of partial sums {5,} of the infinite series diverges, but does not g '0 + © or —, then the sequence {s,} is said to oscillate, Then we say that the series Sy oscillatory series 4 Examples 11.1 I) 14242444... ¢ get ptt @) I-141-14-. 2.2.4 General Properties of Series nes ai 1, The convergence or divergence of an infinite series is unaffected by addition or removal of number of terms, ‘The convergence or divergence of an infinite series is unaffected when each term of the multiplied by a non-zero number. 2. » 16 Yu, and ¥ », are convergent series with sums a and b respectively then for any paitd numbers d and pt, the series (Au, +pv,] converges with sum Aat po. cot 7 2.3 SERIES OF POSITIVE TERMS als The discussion of the convergence of any type of series of real numbers depend upon the Positive terms. So, we shall discuss in detail the series with Positive terms, Definition2.12 A series ¥ u, a» Where u, >0_Y NEN, is called a series of positive terms. a A series of positive terms can cither converge or diverge to, It can never oscillate. 2.3.1 Necessary Condition for Convergence of a Serles Thoorem21 Ifthe series of postive terms Bu, is convergent, then tim, rt Scanned with CamScannerSequences and Series @ 2.11 Proof Let Su, tute tu) tt, then re Sa 5, = Sea Since Lu, is convergent, lims, , where / is finite. limu, = lim(s, ~s,.,)=lims, -lims,_, =/-1=0 This is only a necessary condition, but not sufficient. Thats if imu, = 0, we cannot say the series is convergent, For example: the series 1 1 _ 1 a as Lette ae is divergent, but lim, = lim— Basi ae ee een *If limu, #0, then the series Yu, is not convergent. a a at 2.3.2. Test for Convergence of Positive Term Series The definition of convergence of a series depends on the limit of the sequence of partial sum {s,}. But in practice it will be difficult to find,s, in many cases. i So, it is necessary to device methods by which we can decide the convergence or divergence of a series without finding the partial sum s,. ‘A standard technique used in studying convergence of positive term series is comparison test. The given series Eu, is compared with a known series Dv, which is known as auxiliary series. 2.3.3 Comparison Tests 1. If Zu, and Lv, are positive term series such that u,Scv, VneN, for some positive constant c, then Eu, is convergent if Lv, is convergent. Ifu,2v, ¥ nN and if Ly, is divergent, then Lu, is divergent. 2. Limit form: Let Zu, and ¥v, be two positive term series such that li Then Lu, and Ev, behave alike. If Xv, converges, then Zu, converges and if Dv, diverges, then Zu, diverges. Here we compare Eu, with v,. Proof Given Du, and Lv, are series of positive terms and lim. ., we have 1 > 0. Since “s. 5 9 for all n= 1, 2,3, ¥, Scanned with CamScanner2.12 m _ Engineering Mathematics Choose €> 0, such that /~e>0, Then by the definition of limit, there exists a positive integer n, such that , ~ec tice [e lx-al ny, = Lu, 20 Hence, Dw, is convergent. Case (ji): If Zy, is divergent, then Dv, ->e0 as nye Consider (I-ey, U-e)Ly, > Lu, as neo Lu, is divergent, Note 1. If1=0, then Zu, is convergent if Ev, is convergent. 2. flo, then Du, is divergent if Ey, is divergent. 3. Imorderto discuss the convergence of S.u, by comparison test, we consider Sv, whose is known already. ‘Two standard series used for ‘comparison are the following. (@ The geometric series with positive terms @-+ar-+ar? +.--, where a>0 and r>0 It converges if 0< <1 and diverges if r>1, Jt tL ee It converges if p > 1 and diverges if p <1. The p-series is also known as harmonic series of order p. " Inmany problems, the auxiliary series is chosen as the p-series for particular values 0 ii) The p-series is 7 SA, where p> 0, i Scanned with CamScannerSequences andSerles @ 2.13 For choosing the auxiliary series we write 1, in the form, (2) then decide v,, a fi i 1 1 Forexample i'w, = (4) ‘Then we take'v, =-L,p>0. a” " WORKED EXAMPLES EXAMPLE 1 ‘Test the convergence of the series —!_4_3_4_5 4. 12:3°23-4 3-46 Solution. Let the given series be Du,. Se "7123" 23-4 3-4-6 The numerators 1,3, 5,...are in A.P.So, the n™ term is 1+ (m—1)2 = 2n—I]. In the denominator, first factors are 1, 2, 3,... and the n term is n, the second factors are 2, 3, 4,... and the n™ term is (n + 1) and the third factors are 3, 4, 5,... and the n term is m +2. 1 2 2n 3) [= Sone] n by comparison test Lu, and Dv, behave alike. But Ey, = EL is convergent, since p=2> 1 in p-series cs Hence, Zu, is convergent. EXAMPLE 2 Test the convergence of 3 (cee), mow Solution. Let the given series be Du,. “ Lu, > ; ‘at Scanned with CamScanner2.14 m Engineering Mathematics ‘Then the n™ termis 1, = dn ade h ‘Tuke », leaks) fete) v, In (1+0)(1+0)=1 (#0) [ | is divergent, since p =} <1 inp-series. But Dy, =I olan Hence, Su, is divergent. EXAMPLE 3 Discuss the convergence of ¥: (t+ —n). Fy Solution. Lu, => (Vien -n) t us Then the 2" term is 1 = ie nantes naa{(ve4) -1] " n For large values of 1, + <1, so expanding by binomial series, we get, 7 Let the given series be Du, . Scanned with CamScannerSequences and Series | 2.28 1 Take ¥, =—> <. by comparison test, Zu, and v, behave alike. Ll, A But Lv, = E— is convergent, since p =2> | in p-series. n Hence, Lu, is convergent. ‘EXAMPLE 4 Test the convergence of Y.(Vn'+1 — vn*=1). P= Solution. Let the given series be Zu. «+ Ea, = (Vat +1-vn=1) = ‘Then the n term is u = Vn* +1—vn" =1 (atari —Vat 4) (Sn e1 4 Vn =1) Vit ++ vn*=1 ni +1-(n'-1) Vint +14Vn'=1 " jim“ = lim ———— ‘e hota fs Scanned with CamScanner2:16 Engineering Mathotatios + by comparison test, Su, and Sv, bchave alike, But Sv, = 2-4, i converyent, since ps2 1 in peseries, r; Hence, 3:1, is convergent, EXAMPLE S Discuss the convergence of $ (aa ~ a Fr ” Solution, Let the given series be Ey. Says py = A) wl ” vn ‘Then the an term is 4,2 eal _ tT ~SaN dai + Va) a (Vn tan) ppd ED a (Jn dn) aa J 4 ot Witlrda) wal fea) ” + by comparison test Eu, and Sv, behave alike, Buty, =E iscomergen it p+ >1 = 17> anderen pt wt nt Lg sol Hence, Xu, is convergent if p > ny and divergent if ps a Scanned with CamScannerSequences and Series @ 2.17 EXAMPLE 6 Discuss the convergence of > solution. oT Let the given series be Du,. ~ Dae A x a+ «1 by comparison test, Du, and Ly, behave alike. ( ) is a geometric series with common ratio r= . Ly, is divergent. Hence, Du, is divergent. EXERCISE 2.2 TTest the convergence of the following series: ee - 12°23 3-4 < 1 2 3 , p>0,q>0 4, 4 tt a > Sinpan ed 142" 2? 142? Scanned with CamScanner2.18 Engineering Mathematics 1. Test the convergence of the series 5 i | x Vntdnel 12, Test the convergence of (Jn? In), 13, ‘Test the convergence of the series Y2—1_, V3-1., V4-1 Par Fa SP H1 e100, 14. Test the convergence of the seties 243, 4 a. 15, Test the convergence of F | in, ‘nal n ANSWERS TO EXERCISE 2.2 1. Convergent 2. Divergent 3. Convergent itp +q > 1 and divergent ifp +q <1 4, Divergent 5. Convergent 6. Divergent 7. Convergent if p< 3 and divergent if p> 3 8. Divergent 9. Divergent 10. Convergent 11. Divergent Aes 12. Divergent 13, Convergent 14. Convergent if p > 2 and divergent if p <2 15. Convergent 2.3.4 De’ Alembert’s Ratio Test Let y u, be a series of positive terms such that tim =I. Then the series Du, is a me ' m= et 1 > 1, divergent if (<1 and the test fails to give a definte result if = 1, That is Zu, may converge or diverge if / ='1. Scanned with CamScannerSequencos and Sores 2,19 Proof Ree eres it 11 Eu, isa series of positive terms and tint oJ, Give tim! sinven, 20 WHEN > 0 => 1>0, , lo Since im—- by definition of limit, given > 0, there exists a positive integer , such that ey, nat 4, nce Vn2n ae u 2 -e1: Choose >0 such that /~e> 1, then [-ectel+e WV n2m. [forthise, m ism] Mas [-e< eV n2m Consider u, nt ae sie Vnzm i, = not Replace n by m, m+ 1, m+ 2, oom 15 we get t 4 Mn joe, Heh >I-e, Monat Ming Multiplying all these inequalities, we get a [nm factors) 4 >(J-e)(t-e)-(U-e) uy, Maer Manse Mines Un Mo (Je) Vazmtl 1, = ty 5 U8 y nz ml u, @-e)" Scanned with CamScanner2.20 @ Engineering Mathematics = 1 Go ou, Vazmel => 1 <-O Vaeme+l > Eu, 1 = er ie convergesasn—>e> +. Su, is convergent if > 1 y (ii) Let?<1: Choose €>0 such that /+e<1. Then there exists a positive integer & such that a =e fight 1 oe = FO) Tee tao le Ss 4 > a (HO)! Te 1 = Eu, >m(to Bo oe Scanned with CamScannerSoquencos and Sores @ 2.24 But Y— ue ina geomeltlc artes with common ratio pa oot [vttech—eot Ie, ee va the series SL is divergent, (+e) Hence, Lu, is divergent if! <1, iy Let = 1. Then Ex, may converge ordiverge, Consider Zu, = 4. i ae Wehave tim = tim 2+ in( +) =I fo ater, But 5 is divergent, sinee p = 1 in p-se n Now consider the series Su, = 5% i a ty a (tl)? le sim lim: lim =lin( + But 5 is convergent, since p=2>1 in p-seies. So, when lim =1, the test fils to give definite answer as Ey, may be convergent or divergent. be ‘ “1 Note Sometimes this test is stated as below. 21, then Zu, is convergent if < 1 and divergent i> 1. The fest fails if = WORKED EXAMPLES ‘EXAMPLE 1 Test the convergence of the series Solution. nt" Let the given series be Zu, = (nt 12" and ta = GD ‘Then Scanned with CamScanner2.22 @ Engineering Mathematics uy _ nia" (n +1"! tayo (nea (ot _ (nt d(n+" n(n - (nel) Hina tin 21+ ‘) al fe 21 in p-series. But Lv 1 “Yay, is convergent if ses is convergent if 1. Hence, the given ser EXAMPLE 3 ‘Test for the convergence of the serles 1, tig wl x! Joy te4 a taet mi 32 43 4 Solution. Let the given series be Du, Su a nT wa sSNA x et wt 0m Then = Hat = (atin ™ = GaltDNatl -Ginatt 2 _ ert (nant na [ntl iy) feel oe nx ae thy raetctactershet ¥ Su, is convergen .. by De’ Alembert’s ratio test, and divergentif_-<1 = >l > x<-lorx>l ¥ If x0, the series is convergent trivially et}, the test fails to give conclusion thes x1 * x Scanned with CamScanner2.24 Enginooring Mathematics A a tl I 1 In this ease, the series in 4 oy Lge wi 3a” 43 ! 1 J os uM, G+pdn ‘[u-t] Take v, F n. w ne So, by comparison test Su, and Ev, behave alike. But Ev, = 5+ ts convergent, snes pr 3 >| in p-scrics . ++ Zu, is convergent if x? = 1, When x= 0 the series is trivially convergent Hence, the given series is convergent if -1$ x S1 and is divergent if x<—lorx>1. EXAMPLE 4 Discuss the convergence of the series 1 1 1 1 mt et et et vee, 7 i tee ieoe' iat teat for positive values of x pa Solution, Fe! Let the given series be Zu, in 1 L —+— 5+ + eT tien ne ae Then: 34. le Tem T+@+ bx Le (ne Dx™! ; User l+nx" 1+ (n+ Dx" oy = [Dividing Nr and Dr by nx*] bgp Scanned with CamScannerSeuercesardSeres © 225 a which is divergent. 2 Zu, isdivermentif0 I and divergent if 0o 2 2 ae Scanned with CamScanneray" fog? | pigege - Sia vest w Waa haan’ 4 toy ahed aly! ay? . wo, | ng ot eo) a Lad Waa has xt i Sede, peso gun, gd “Dy "nas haere Yael v¥et Viet Wa 1 R pies a re tf meen he var wa sar : ! Ao TO Wear DQarh (ht DDE) AS, Prove thy cries HEE WONG D | rove thatthe series 1 bal (edhe converges if.) > a> 0 andl diverges ita b> 0, we 12. Test the convergence of the series shown below oa, p> O and x >0, SF oe by ral Sobers by ratio test, y x>Oand yal ANSWERS TO EXERCISE 2,3 1. Convergent for all p > 0 2. Divergent 3. Conve 4. Convergent if'0 1; divergent ifs 0 8. Diverg 9. Convergent 0 11, Convergent if 0 1 12. Convergent 13, Cor 14, Convergent if 0 Ound ify L,p> 1. Mis divergent ity > 1, p > Oa 18. Convergent 19, ‘The series Zu, is convergent ify |. IY isaseries of posit wt ‘The test fails to give a definite conclusion if / “x by the definition of limit, given e > 0, there exists an integer n, such that 4 1 ed ce tae = I-e

0 such that +e <1. Then there exists a positive m such that 1 I-ecubcl+e Vn2m = (=e) 1. Choose € > 0 such that /-€ >1. Then there exists a positive integer & such that , 1 I-ek Consider (Ie) (I-e) Vn2k Du, > L(l-e)" But L(/—e)" is a geometric series with common ratio r=/—-€> 1 L(l-e)" is divergent. Hence, Lu, is divergent by comparison test. Scanned with CamScanner2.28 m Enainooring Mathomaticn (i) Ue 1, then t est {ils to give n definite conclusion, \ ; 1 1 a Consider ¥! andy avo find that dima! = in( 4) =f and lima, = tin( 4) q ii a not eve GD nore nae ol . = But y is divergent and yt is convergent, w ++ Zu, may be convengent or divergent if | Note 1, Root test is more general or stronger than the ratio test, because there are cases where test fails but root test gives definite conclusion, 2. The root test is used when the general term x, contains index interms ofn. WORKED EXAMPLES EXAMPLE 1 Test the convergence of x(t + +) : a Solution. Let the given series be Zu,. i Eu, =3(142) £ = ' a! ° watt & f d eelfiet) | alia Then ne(t4) = (4) -l(-) | [i+4] 1 NY" 3 limut = tin(1+4) fim ay = Tien ws by Cauchy’s root test, Lu, is convergent, EXAMPLE 2 = 1 ‘Test the convergence of > (1 + Th Solution. Bu, 23 (1+ Let the given series be LM. > Scanned with CamScannerSequences and Series @ 2.29 :. by Cauchy's root test, Lu, is convergent. EXAMPLE 3 Test the convergence of the series given below 2 a (@ 3) ( v) *F7 2, Solution. Let the given series be Du,. Ll FT ET ot . ne n Le STS) floTe oT] eso a) tees ot} Od) ot] sea “+ by Cauchy's root test, 3, is convergent, [s 2 1] Scanned with CamScanner2.30 @ Engineering Mathematics EXAMPLE 4 Discuss the nature of the series 2 . ztie +(3) xe 3) x! tece for x > 0 4 5 Solution, Let Zu, be the series, omitting the first term, a 2 ya “Q)e x aye + ae 5 Then a a i ne2. n+2 uy, 1+= n 1 \ It n Jimug = lim x =x n st wt (4) £ ry (224), n =(24)),s— (=): n+2. 2 3 :. by Cauchy's root test, Eu, converges if.x <1 and diverges ifx > 1. Ifx=1, the test fails to give a definite conclusion, In this case, u, -(24) —— n+2 (# n+l £. when x= 1, the series is divergent by theorem 2.1, page 2.10, Hence, the series is convergent if x | and divergent if <1. The test fails to give a definite result if /= 1. We first state the theorem of comparison of ratios of two series. a 7 a? If Zu, and Ey, are two series of positive terms and if “sit < “esl for all values of m€ N, ta Uy Yn : 4 %, Lu, is convergent if Ly, is convergent, whereas if Moss 5 Yast for all values of m, then Eu, is ey, divergent if Dy, is divergent. oe Proof i ; Given Su, is a series of positive terms, we compare with 5 v,, where v, 0 We know that S.v, is convergent if p > 1 and hence, Zu, will be convergent if Ev, is come and if VneN and if p>l WneN and if p>1 But Scanned with CamScannerBut Lv, is convergent if p>. +. Zu, is convergent if im: (2 -1) > p> land divergent if tina Hg ty tat Alea ft The test fails iti 2-1) =1 ne Vo ofp sun WORKED EXAMPLES EXAMPLE 1 Test the convergence of the series Solution, Let Du, be the series, omitting the first term, ant " Then oj plese Onc Rill At 9 46e-2n (2nt1) 1.3-5(2n=DQn+1) x8" i and Ui? oe oD) ned) net 9g. 6-dn(2n# 2) (2n+3) n=l) 281 2-4-6 yn(Qn+2\2n¢3)_ 1 TF 5an-bene) ( 4 ees). 0 5? ie (244) : Scanned with CamScanner2.38 ™ Engineering Mathematics ay ae lim—- = lim: mt ( +. by the De’ Alembert’s ratio test, Now pel = x%=1 = x= x Ifx= 1, the test fails to give a definite conclusion. In this case, we use Raabe’s test. When x= 1, the series is | Qn+2)(2n+3)=2n+ 1 ary _ 4? +10n+6=(4n? + 4n+1) nt S a (n+l? Qn+1P 2 5) 5 60-8) _ m(o45 OF > (n+l) «(0 1 «aby Raabe's test Eu, is convergent if Hence, the given'series is convergent if 0 1. Scanned with CamScannerSequences.and Series 2.39 EXAMPLE2 i ‘Test the convergence or divergence of i 2 xt pat xo 3-4.5.6 eae the first term, let the given series be Zu, 2 2g 22.4? 6% } oe ty 6 Sea fad 34-56 3-4:5-6-7-8 ee On) On 3? até | 834 5-6---(nt I)On-+ 2(2n+3)(2n+4) uy Qn43)Qn44)) 1, ty Ont2 and. 2 by De’ Alembert’s ratio test, Zu, is convergent if 1 ote (Sp-bex: x As Ifx2 = 1, the test fails, so we use Raabe's test. Inthis case, (2n+#3)(2n+4) (2n+2)° Si _ Qn+3)2n+4)-Qn+2? 5 a (2n+2)? An? +14n+12—(4n?-+8n +4) __6n+8 ST Qne2e 0. Qnty nis [From (1)] Scanned with CamScanner2.40 @ Engineering Mathematics veloxeatl. by Raabe's test Du, is convergent if. : when x = 0; it is trivially convergent] «the given series is convergent if -11. EXAMPLE 3 ie Test the convergence of the series Le 1? ——, gar Solution. The given series be Eu, “+ | Duy = 00 Spi ani Then (nt bypxt? _ [eI _ Moet =" Dnt DI! (2n+2)! a_i Qn+2)! (2m! [+f x? Mya _ (al? Qn)!Qn+DQn+2) “ Qn) (@lP@rtly _QntD204) 1 og (a+1y ft ty 2Q2n41) 1 ha aL \ Scanned with CamScannervu by ato test, Ey, is convergent 4 > ay xed mp oxta4c0, = ~2er<2, 440 F : : When x=, trivially the series is convergent and divergent ifx<~2 or x>2 { nay 2 = 4, the test fails. So, we use Raabe’s test, qh ARES When tat => x’ x Inthis case, from (1), we get wh ] Peat ia 2an+l) 1 (2n+1), nt 14 Und, tng! Bn 1~2(n4N) 2+), Ane lyr) ib af ty ala tine( 2-1} tin 7p T ea a{i+2) n, 4 aie” -. by Raabe’s test, Eu, is divergent if x <. the given series is convergent if -2<.x<2'and divergent if x<-2 or x22 EXERCISE 2.5 Test the convergence of the following series. wii Scanned with CamScanner246 Engineering Mathematics He eee 2n 8 nt lim nlog, ae tin(—142 : ma 2B a ast by logarithmic test, Xu, is divergent if x= the given series is Convergent if 0 4. 2.4 ALTERNATING SERIES Definition2.13 A series of the fo nating series, a yt toe here uy, >0 WY me N iscalleday Examples a Wied - is an alternating series, . 2 @) tor is an alternating series, at at 2° ga Thatis a = - isan alternating series, bécause it is +1 24 2 ava 2.4.1 Leibnitz's Test Statement: If the alternating series hota Fy “us toe is such that Gi) ty 0 Vn and (i) 4, 2u, 2m 2u,2-u, 20,2, and i) Hime Consider the even partial sum s,,. Soy = (ly “t+ (Us Wy) Fo + (a 4) 2 Wag t Mag) =1,,20 Vn, Since Hagar 2 Magy Hager “Han a Scanned with CamScanner‘Sequences and Series a 2.47 Say 2 Sane Vn * 05, $5485, SS, S> tes write cowe cat . gp == (1) =) ~ Wager Hae)“ difference in the brackets are non-negative anid u,, >0 Ya very Set, Wn ; the sequence of partial seems {s,,} is increasing and bounded above. eee, itis convergent 3 lims,, = Now compute the limit of odd partial sum. Now Soaet = Sant Mant Himsa = His, ta) Since imu, =0 and lims,, =, lims, =, we have lims,,., =/+0=1. since both even and odd partial sums converge to, we have lim, = 1, and so the series is convergent When lima, #0, Tims, # Timsq.1 «the given series is oscillatory. Tus inan alternating series ifthe terms are decreasing with lim, =0, then itis convergent. Ml WORKED EXAMPLES ‘EXAMPLE 1 1 i 1 Test the convergence of the series 1 ——= + —=——R— ts. 22 33 Ava Solution. The given series 1-1 1... is an alternating series with a2 ws ava 1 1 ee mn nln] We know that (n+IVnel> an > 1 1 ee Say sg VAN. (a+ DWntl aa Scanned with CamScanner2.48 @ Engineering Mathematics ic., the terms are decreasing. Now lima, = lim: me ed Hence, by Leibnitz’ test, the given alternating series is convergent. EXAMPLE 2 Test the convergence of the series (ue Solution. : The given series (—1)" .»- {gan altornating series with n+l ntl we nel n=l _ (a +D(Qn=1)—n2n +1) "QneD2n=D dn? en 1-(2n? Fn), ei 4n?=1 uy 00] -: by Leibnitz’s test, the given series is convergent. EXAMPLE 4 Test the convergence of (ew (tay } ot (+n) , Solution. Lee tale ees + The given'seriés is 27 4 #2143" 142% 149 1eP Le? 143 142° 143) Consider, the series is 142 4243 Lea 15 TP LED 1S ea 145 It is an alternating series with Le STe ay | +n LEAL)? 4307 43042 me+2n+2 L+n? w+3n43n42 140 i (14m )(n? +2n+2)— (ta? lw? +3n? +3n+2) = (em (+ 3n? +3n+2) Qn Ont ni + Qik Qn? (+ 3n? 43n+-24 n* 430" +30? +20") % (EWP? £30? + 3n#2) 4 tana tm) og [+ mis positive] "(4 PP + 3n? +3042) ty

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