1014convtests - NOTES ON CONVERGENCE TESTS FDR INFINITE...

Info iconThis preview shows pages 1–10. Sign up to view the full content.

View Full Document Right Arrow Icon
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 2
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 4
Background image of page 5

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 6
Background image of page 7

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 8
Background image of page 9

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 10
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: NOTES ON CONVERGENCE TESTS FDR INFINITE SERIES rill infinite set of numbers. {In} may either have a well-defined finite sum or not1 in eonti'ast to a finite set. of numbers which always has a well—defined linite surn. Sometimes it is cont-'enient to start the labelling of the set. with a first integer, N5, (lillereni from i tag, on 2 life. — 2‘] is undefined for n. : 2, so we would typically only try to add up the on starting at some point rt 3 3]. The general Form of the sequence would then he {fm n. 2 at}. [F the infinite set of numbers {In} DOES have a. finite sure we call the infinite series fit-J + flea“ + - ~- E ZEN: n convergent. II' it does not have a well—defined sLI1n._ we call the in finite series dieeryent. Convergence {the esistenee of a well—defined sum} is defined using the Following limiting procedure: ll] From the infinite set. {fin a = NR1 - - - . Form the related infinite sequence ol‘ partial sales {335-}! where 53; is delined as the sun] of all terms from the start (Ne) up to and including the .-""r"”‘ term, 5'»: = fw, 'l' f.-v,+| ‘l' ' ' ‘ ‘l‘ Is- tii} If tinting-writ exists and is eqilal to .3, we say that BEEN: f" is convergent and eqind tn 5'. [If the limit does not exist [either by diverging to :izoo OR by oscillating and failing to settle. down to one fixtitl value] we say that the infinite series Zia-i f“ is divergent] S UME GENERAL RESULTS Xti'lL-t'l'lflr‘v': Sinee different infinite series can start at different points [different values for the label, at” of the “first” term in the series}, let’s use the notation E“ f" to inrlieate a generic infinite series which starts at sorae NH and involves an infinite number of terms. ii'e have seen already.’ the following general results: {1] If i-iirr.,_.mf,, 75 F], then E” f” is divergent. NOTE: as already discussed in eiess this UNL l” teiis you that if the iimit of tire f” as n. —> so is NO T zero. then the series is for sure dinereenii. It is stiii perfectly pessiirie to have a. divergent series whose terms go to zero as a mi- oe {the firmnenie series, :1 1 in. is an. erenipl'e of seen. :1 divergent series). {2'} If E” f” is convergent and equal to S, then En efn is also convergent and equal to CS, For any constant c. Similarly. if E" f” is divergent and c sé fl is any HUN—ZERO constant, Z“ all“ is also divergent. {3} If E" f" = 3m and Zn 5.1“ = 5”] are two convergent. series1 the two series with terms I” —-.- on and f" — 5i“ are also hoth convergent, with E" [fn :l: on] = 5”} i 3M. TESTS OF CONVERGENCE FDR POSITIVE CIR NON—NEGATIVE SERIES :‘i series E” f“ is called pesitit‘e if f” :3 [l for all n. It is called eeentuntty positive if f“ :- ti for all n greater than some fixed value. it series is, similarly, called hen-negative nr eventuality Iio-h—riennti-ee ill either all the f" l“_-" t], or f,, E [l for all n. greater than some iii-Led value. ‘r‘i’e new study some tests that help us determine whether a given IIDfl-Iiflgfltih't‘. series is coin-'ergeut or dirergent. 3i nee the sum of any finite number of terms at the start of a series is finite, the enu‘i'ergenee or divergence of a series can alu-‘ays be determined hy ignoring any finite nnlnher of terms at the lieginniny,r of the series. Thus, the conditions for an eventually non—negative series to converge heceine equiralent to the conditions for a. non—negative series in ennverin: it we simply neglect the finite number of terms that occur hefnre the eventually non—negative series settles down to having only non-negative terms. For this reason I will state the convergence tests below only For the case that the entire series is nun—negative. This is only to avoid a lot of extra typing that 1ironld he required to state the general ease. You should understand that all results stated below alsn hold if “unu—negative" is replaced he “eventually unnvnegntiee" and fer “positive” is replaeed hy “m—‘entnally positive". {1} The Bounded Positive Sum Test: Since, for a non-negative series f,, 3 ti, and S's-44 : 3.1- + f,,+], we see that, for such a series, tier 3 3s for all N, and hence that the setpimice nt partial sums, {5's}, is non-decreasing. From an earlier result on hounded neu— derreasing sequences we thus had the result: A nee—iieeeiine series is eeiieeipeiit if it enii tIe shnimi that the serjruen-ee sf its pertiei sums is heiiuded fi'eiii she-ire. {This result is mostly useful the proving that ether tests 1.rhich are generally more helpful in practical applications are true, but snmetiines it. ean he applied to specifie cases as well — we will see an example of this type in class] {2} The Integral Test: This is a generalization of the method we used iu class to show no rr—I that the p—series Z 1 in.“ is convergent it p :5 l and divergent it p it 1. The hash: result is as tellnws: The integral Test: it" 2:32, _,I",l is it pesitiee series for which I" IS A NON-INCREASING FUNCTNJN 0F ii, and if one can final it POSITIVE, NON—INCREASING FUNG'TIUN f{;i:} such that f“ = flit}, then. in) if the improper inteyint if” f [IJJIII is convergent, the series 2:] f,, is also convenient, and I32! {h} if the iniprnpci' iiiteyrni if: f{s?}it;r: is divergent, the series Er. ,, is else divergent. PHUU‘F UP THE INTEGRAL TEST: The result. can lie unrierstrititl [rem the fellewing figure1 antl is basetl en the observation that, SO LONG AS Its] IS A NON-INCREASING FUNC- TION OF .‘t‘._ the tops of the large rectangles lie above the eurre and the teps of the small rectangles lie below it1 antl henee the area at the set of large reetangles is greater than or equaE tn the area under the curve, which is in turn greater than or equal to the area of the smaller rectangles. The rest. at the argument goes as follows: I- The reetaagles all have width 1. Siaee the small rat-tangles hetween a: = 1 and a: = a have heights ft‘EJJEEL - - - . fin}: Iglf31"‘,ffll. and the earrespending large reetan- gles! similarly, have heights f1,_t-_:_.~ - -,f,2_1. it fulluws that n nai- I ELI was a If If” rtr flier} diverges (to +ee] as a —> 30.. the seeend at" these equations shows that EH21}; diverges tu +0:- also. 0 Similarly. if f1“ dwaI} eeaverges as n —} 00.. then sinee fire} :5 {i implies that If” rtzr f (:1?) d: tilt; the first at these equatiens shares that the sequenee el' partial sums, S“ = I: + 215:2 fin is bounded ahere by f1 + If” the fit}. The series Efl] f" is thus {:[ilit'ergent h}! the Bounded Positive Sum Test. ."'i't.}i!"t:'.' fer the 'ti'ttt'yt'ttt test te he itsrihte tt ts criteth that the {In} satisfy 30 TH ef the ma- rtttterts fitter-we, Mamet-y tttm‘. they tie {eventuality} prisi'tit'e, AND {silent-mam) rtrm—taerems-hty. Students after: feiyet ta cheek the sees-mt at these marh'ttans. Hamlet's-r, sittheirt the second t-eiirhttair. the titfl'I-‘e' of Its] eeriht rise up [these the tap sf the huge reetaiiyte, er {ftp detest heteie the top ef the san reetengte eerr‘espencttng ts twe ectjheent integer setnes sf 3:. If this hegpe'nett, the t‘i‘Hr} tneqnetitz'es nsett ebe‘ee ts pmee the settitttg ef the tntegrnt test wenht nu ts'sger he wettet. SD...DDN1T FORGET TO CHECK FOR BOTH PROPERTIES BEFORE USING THE INTEGRAL TEST. [3) The Cemperisen Test: The basic result is es teliews: The Um'hgrm'strn Test; .5" e J tf gen have n “new” nsn—negettee series. Zn In, entt sen reeegntze that eeeh term sf the new sertes ts 33 the eermsprnnttng term s." eh “std” nenrnegettee seriesJ E" g“, etreee‘g hes-em ts rt-teerge te 3-5,. then the new series etss diverges to 00; {h} if gen heee it “new” 'rtsn-negett'se sertes, E“ f", met een mesgntze thet eeeh term. of the new sertes ts :1} the eerrespentttng term. sf en “std” non—negative sertes, E" gm etreertg hnrnsn ts he eeneergent: then the new series ts etse esneergent. PROOF: i If E“ g" rheerges {tn DO ,1 and f” E- gn 3 E]: If3 and Slat] are the NE" pert-is! sums fer the f“ and g" series1 respeetively, f“ 3 g“ implies thet Sit: E 533'). Then, since SEE-r} diverges te era as N —> 00, se Lie-es 3H3, as ehii need. I If E" g“ eeneerges and fl 5 f” E g”: in this case, f” E g” implies that. SH} 5 Sigh. If the hi]! sum! E“ g" = G then, since th 3' ii, 55:?" s: G and hence SEQ-r] s: G as well. is therefere a hellnrieti erm- tieereesing pesilive series~ nut! heeee eem‘erges by the Eeumieti Pesitive Sum Test~ :—1s eis heed. NGTE.‘ thts meg sf phmstng the resett, 'tehteh ts sentewhet mere eetteeetet then that given tn the test, emphastees the rung tn n'hteh gee weetd eetnettg ge sheet epptgtng the test. ngiir eettg gee ere gteen sente sertes E” f“ :1an rten’t get hes-ts whether at eenserges er rtteeijges. The. Cerhjmrtsen Test nits:sz gen te hnnt fer ensther ser‘tes En gn get: are etreettg femtttsr with the. ene where gen etreertg hittfl‘ttt whether it ts eeneergent er thee-Went} met use geer hnen-‘tee’ge sf tts esneergenee er rtteergenee te finrt set sheet the new ser‘tee. The "henttng" shsetrt he guided hg gen‘r' “QUE-315'” street whether the new sertes ts tthetg ts he eeneergent er rtteergent: if gee. thtnh‘ it might he eeneergent, then Tehet yen went ts test: for is e fe-mstter errnvergej-It series with terms feesntnettg} htgger then these ef the new series; tf gen thtnh tt might be divergent, yen went to feet: for e feiititirir {iterates-st series iehese terms ere {seen— teeiiyj sntettei' then these ef the new series. We will tie seme examples ei' the use of the Cetnnariseu Test en the heard in elass. The Limit. Cemparisen Test: The errliuary eelnpitrisen test ean sometimes allew yen te easin shew that. a given new series either eeIn-‘erges er diverges. but fail in give sueh ini'erniatien [er anether new series whieh leeks very similar to the prerieus ene. Fer L‘KEIIIIIJIIZ‘... if [n — i} 3} I fit. i'er n. :2— 2. Sinee 2:2 lfn is tlirergent (the tail Hi the llarmenie series] the D: H: Cetnparisen Test, therefore, tells us that Z 2 Min. — 1} is alse divergent. iiewerer, if we new eensider the 1very similar series Eff; i 3’ [11+ i) we eannet immediately use the hermeuie series tn peri'errn the eenlperisell oi the Centparisen Test heeunse the terms ititt + i} of the new series are smaller than these ell" The harmenie series. net larger. Being smaller than the terms iii the harIneIlie series it eeuld tIIerei'ere still he pessihie fer them te add up te a iiniie tetal. rl'he Limit Cellipflriseu Test is a test that hesieeiiy allows you tr.) argue eleng the litres ei' “its 'i'i' —.'~ :10. Itin- + 1} gets te hehave mere and mere like 1 f It... se 1 weeid espeet E3; 1 fin + ii to diverge just like 2:; lie. dens“ [and is have this argument he eerreetli. ii again irii-‘eli-‘es you being given a “net?” series Zn f" and, free: the hehavier oi the terms ei this new series, trying in find anntller mere “i'aItIiiiar” series E“ y“ te use ier melting it trelnparisen. 1Ll'niy in this ease the way the eelnparisen is delre is different. The Limit Ceensnisee Test: If Zn in is e. earl-iieyetiee series and E" {in e jmsitiee se— ries {pastime tether thee- j-est e-ee-eeyetiee se that fate” mitt always be defined}. than F tinnhm = L emit-s met is finite THEN {a} if L eé ll}. if guy” is divergent, En f“ is etse divergent; end {it} if Eng" is tfri'tt‘t'tii‘ytfflt: E“ f” is etse eermeryent; I" h j if L = ["1J (i) if Zn ,3!” is eeneeryent: En f“ is etse eerieeryeitt; white {iii| HE” g" is (tit-'er'yent, the test gives me infei'metiert ehettt whether E” f” eent‘eryes er diverges. PROOF: The preef relies on the ertlinary Cemparisen Tlleereni. o The ease L # 1]: Per large eneugh n the rstie jute” must e1-e11tueliy get as elese as we want in L. in partieulsr. we earl ehese to make “elese” mean t = LIE; there is then seen: my such that: fer all n. :> N, L}? 4: fut.ng s: ELIE. er. equivalently. sillee en 3:- CI. {Lfflleu s: f” c: [HLIEMH I’er all n “e Na. -— If Zn 5'“ is convergent. sn is ZHIIHL-Jtfiiym since 3L}? is El eensmnt. The inequality [3 <_Z f“ c: [Sis-“2m” then implies. from the ordinary Comparisnn Test: that E“ f“ is flit-it] ennvergent. m Similarly. if Euler, is dit‘m‘gent, sn is EntLfliyn, sinee Lt? is a constant. The inequality f” 2:- [th 12in” then inlpiiesl item the ordinary Celngiarisen Test, that E" f” is alsn- divergein. a The ease L = [1: The esistenee {ii the limit means that fntgn [a [resitive Innnher} erentnnily can be made as snntli as we want h}! gnng nnt tn large enough n. in particular, we can iinti Nu 511ch that. Ii] 5 filter" s: 1 Fur all n 2:- Nn, Le. U i: In at y“ fer Edi it is iii-"u. if Eng” is ennrergent, this innrnlality. eernhined with the el'dinarj.‘ Comparison Test, imniies that E“ f” is aisn ennvergent. 1! Te see that the test gives nn infernnttinn in the ease that En y“ is divergent and L = D, we need nnljr tn shnw twn examples [if this type, nne where Z” I” is eenrergent. and annthnr where E“ f,, is divergent. Let y” Uri, se that E” y" is the divergent harnlnnie series. — In the iirsl. example eensidel' f“ = iii-stein”. The series 23:2 .1 enn then he essiijr shnwn tn (iii-itrgn using the integral test. Finite that tirnusmfnfg" = tirrin_m¢1ft311[n] = fl se this gives an example where L = f] and Zn f" is divergent. — in the seeend example. consider the eensergent 'p = 2 p—series = Ex E n: I It Eff; Uni”: and 5r” = ifs again. Ones mere ti'rnnsmfflfng t-t-m..._,5¢1fa 2 ti. This ’rhns gives a different. esalnpie where Z” 9‘” is divergent, L = ii is still sere. inn. E" fr, is tsnn-s-n‘gent, nnt divergent. NH TE: The they this test -reert:s is that yes tire FIRST given it “new” series Zn In. Yen THEN twee te finrt e seemni, “ale-re firmitinr” ssriss Z“ In” is use fer the ernnper‘isen. The they gee-e typieeity rte this is, as suggested title-es, by tossing at f“ and seeing; if it heireees titre sernethirn; simpler e") fer terns it. If this simptsr' fern; eerr'espends in e. series Eu 3;" which. yen sirernty tine-m either sen-ssrnss er rtiiiei‘ges, time this presides n Heed “seeenrt series"J fer 11.!1'1’2' in the Limit Cemprn'isen Test. Weill rin- a few examples en the heard in elnss. [5} The Ratio Test: Students usually let-'e this test lieeause it inveli'es a simple I’ermula and {ii} it tleesu’t require them te esereise their iluaginatiens te eerue up with any seeeutl series uet given in the erigiual prehleru [as the tee types ef eeruparisen test aheve tlitlj. III selne eases it is quite useful l'er this reeseu1 but it is impertaut te remember that the test earl he “luem1elusire”. anti serue ef the eases where it IS 1IJCUl1L'1115i‘t'L‘ are precisely these where either the Cemperisen Test or Limit Comparison Test work well. The Rate} Test: If E“ f“ is e gmsitiee series such the! lire.“_,m [JFHHffn] = It then {e} if 1?. <5. 1: E“ f" is entitlement; {tr} if}? :5 1, En f“ diverges te ee; {e} if ft = 1, the test is trimesters-tee file. Elie-re are same eeses where R = 1 met E" f” 2's meeeryeet, ear! steer eeses where R = 1 bet E" f” is {titre-meet, which means that, if yes rifll‘l‘tp'ltr'ifi R. = hire:r,,_,.:,c [an fin] sen! fies! fillet. tt’s eerie! to 1r yes. esee ea ides. aflrether Zn In is eeeeeiyeet m' rl-teerjeeat mitt! yea perfume some ether test PROOF: The preef werks by using the geemetrie series as the “settean series” in the cem- pm‘isen teen. e The eese H <1 1: — Siuee firenemfibflfia = R s: 11 we ean make fluff” as Close. te R as we went by eheesiug :rr large enough. In pel'tii’flllsr1 for any 1‘ between It and l, R c: r s: l, we ran iiml l"pr large eueugh se that {l E ffl+1ffn «:1: r s: R for all r: :5 Na. — This means that [er all e :5 N, L1H a: rfn. Thus1 few-1 s: i'ftrfl. fire-n+2 :1: I. ffi'U-I-I '5: 1"? fear: I It?f1~'l"[|+k c: ?I.|r:\"u+l.‘—l {bl ' ' ' i: I“: f.“|.'u- '- The sum of the tell of the sequenee thus satisfies '3‘; 13:? I]: -'-' _ .k 2 INU+I¢ ‘5 Z fen 5'" — far-1.21 - L'ZU R=fl 5:2” Sines U s: r e: 1 the series Sin 1"“ is a eem-‘ergeut geeuietrie series: and therefore the origins! series een‘eerges in}: the Hermder’l l’esitlve Series Test, as claimed. a The ease R 2:» 1: — Ev an argument similar to that above! we can choose an 1" .‘illtiil that 1 it 'i' <1: H Hilf] find an NU nueh thnt frvu+k I? 1"]: iv” for all n 3* Nu. — 'i'he Lei] of the original series; then, similarly]. Eintisliefi '3‘: 32' J: E fen—vit- 3’ fivu Z? a i=u i-=n — The seriefi 23:“ :J‘ on the HHS is now a divergent geometrie eerie}; {Hinee r 1% 1} and home the originni series is alt-zen divergent by the Comperieon Test. The eese i? = i: — t-‘v'e knew that the hnrmenie HI-iFiilfi 2:1 ,. with f” = Me is divergent. For thie ease iireflhmfnflff” = emmmei + life : 1 so this it-i FLII example of n tliti‘ifl where R = 1 anti the corresponding series is divergent. — We know tiliii. the series 22:; In with f“ = 1 Keri: being a p-series with p = 2, is convergent. For this ease i-immmffllffn : iiern+xiin+ light." = i so this is a second exempie for which if = l, hut where the eori'eseomlhig series; ie convergent. — Since these two example-3 show that R = I earn eeri'efipoml to either a {:onvergent or divergent eeriei;E we see that finding iimnflmfibH ff“ = 1 tells: us nothing nhont whether 2,, f" is eonvereent or {liverjjrent~ an elninled. gummoni NOTES ON CONVERGENCE TESTS FDR. NON-NEGATIVE INFINITE SE- RIES SDh’IE GENERAL RESULTS We have seen already the following general results: [1} If iimnnnnfn 5&- I]. then Zn In is divergent. NOTE: This does not worlr "the other way around”: thore are divergent series Enfn for which lirnnnnIn fn = I] i" the harmonic series, :1 lfn is an example of such a divergent series). [2} If Zn In is convergent and equal to S. then En sin is also convergent and equal to c3, for any; constant c. Similarly. if Zn In is divergent and c 7‘: II] is any NON-ZEN} constant. Zn c In is also divergent. {3) If Z" in = 3L“ and En gn = Eli” are two convergent series. the two series with terms fn + gn and In — gn are also both convergent, with En [fn d: gnl = 3”] :l: Slfil. TESTS OF CONVERGENCE FUR POSITIVE DH. NUN-NEGATIVE SERIES A series Zn In is called positive it fn :5 t} [or all n. It is called eventually positive if f... .‘e U for all n Igreater than some fixed value. :1 series is. similarly. called non—negative or eventually non-negative if either all the fn 3: U. or fn 3 CI for all n greater than some fitted value. The lolloiving tests are useful for investigating whether a given {eventually} non-negative series converges or divergences. The proofs will he done on the board. NOTE: Since the sum of any finite number of terms at the start of a series is finite. the convergence or divergence of a series can always be determined by ignoring any finite number of terms at the beginning of the series. {1] The Bounded Positive Sum Test: A non-negative series is convergent if it can be shown that the sequence of its partial some is bounded from. above. {2} The Integral Test: JP :31. in is a positive series JFor which fn i3 A NON—iNUREASING FUNCTION OF n. and if one can find a POSITIVE. NON-INCREASING FUNCTION flit} such that in = flail: the“ {a} if the improper integral I 1°” fields is convergent. the series Ea"=L In is also convergent. and {b} if the improper integral If” fields is divergent. the series 2:; fn is also divergent. {3] The Comparison Test: (a) if you have a "new" positive series, Zn In, and can recognize that each term of the new series is 3 the corresponding term of an "old" positive series. En gn. already known to diverge to no. then the new series also diverges to on; {b} if you have a “new” positive series. En In, and can recognize that each term of the new series is E the corresponding term of an “old” positive series, En gn. already known to be convergent. then the new series is also convergent. The Limit Comparieon Test: If E“ f“ is a non-negative series and En gTl a positive eeriee {positive rather than just nonvnegative on that furrgn anti attvage be defined), then IF timnem = L exits and is finite THEN to} if L 5E [1. ii} if Eng,1 ie divergent, EMF" is also divergent: and {it} if Eng“ is convergent. Zn In ie ateo convergent; {in} if L = [1. (i) if En g“ is convergent. En fn is also convergent; while {it} if En gFl is divergent. the test giver no information about ivhether E", f" converges or diverges. [5] The Ratio Test: HE“ In is a positive series such that tirnnem [ffilffn] = R then {a} ifR «c: 1. Zn In is convergent; {h} if}? :r 1, Zn In diverges to on; to} if R = l. the test ie ineonetvsive (i. e. there are some eaeee where R = l and Zn In to convergent. and other (even where R = 1 bet E” f” is divergent. which mettne that, if you rsorngnte R = tlttfln_.-g_ Hue] ff”; and find that it re egnot to 1. you have no idea whether in In In; eonvergent or divergent ontit goo perform eorne other test}. ...
View Full Document

Page1 / 10

1014convtests - NOTES ON CONVERGENCE TESTS FDR INFINITE...

This preview shows document pages 1 - 10. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online