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Unformatted text preview: NOTES ON CONVERGENCE TESTS FDR INFINITE SERIES rill infinite set of numbers. {In} may either have a welldeﬁned 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 fitJ + 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 tintingwrit 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 Ziai f“ is divergent] S UME GENERAL RESULTS Xti'lLt'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 iiirr.,_.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 hennegative nr
eventuality Iioh—rienntiee ill either all the f" l“_" t], or f,, E [l for all n. greater than some iiiLed
value. ‘r‘i’e new study some tests that help us determine whether a given IIDﬂIiﬂgﬂtih'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 nonnegative 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 nonnegative series f,, 3 ti, and
S's44 : 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 nondecreasing. 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 serjruenee sf its pertiei sums is heiiuded fi'eiii sheire. {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 NONINCREASING
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 NONINCREASING 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 rattangles 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 fitterwe, Mamety tttm‘. they tie {eventuality} prisi'tit'e, AND {silentmam) rtrm—taeremshty.
Students after: feiyet ta cheek the seesmt at these marh'ttans. Hamlet'sr, sittheirt the second teiirhttair. 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
hesem ts rtteerge te 35,. then the new series etss diverges to 00; {h} if gen heee it “new” 'rtsnnegett'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" pertis! sums fer the f“ and g" series1 respeetively, f“ 3 g“
implies thet Sit: E 533'). Then, since SEEr} diverges te era as N —> 00, se Liees 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 SEQr] 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 hests 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 theeWent} 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' “QUE315'” 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 femstter errnvergejIt 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 {iteratesst 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. earliieyetiee series and E" {in e jmsitiee se—
ries {pastime tether thee jest eeeeeyetiee se that fate” mitt always be deﬁned}. than F
tinnhm = L emits 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 e1e11tueliy 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 ZHIIHLJtﬁiym since 3L}? is El eensmnt. The inequality
[3 <_Z f“ c: [Sis“2m” then implies. from the ordinary Comparisnn Test: that E“ f“ is ﬂitit] 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“ = iiistein”. The series 23:2 .1 enn then he essiijr shnwn tn (iiiitrgn 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 ttm..._,5¢1fa 2 ti. This ’rhns gives a different. esalnpie where Z” 9‘” is divergent, L = ii is still sere. inn. E" fr, is tsnnsn‘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, “alere ﬁrmitinr” ssriss Z“ In” is use fer the ernnper‘isen. The
they geee typieeity rte this is, as suggested titlees, 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 tinem either senssrnss 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 trimesterstee file. Eliere are same eeses where R = 1 met E" f”
2's meeeryeet, ear! steer eeses where R = 1 bet E" f” is {titremeet, 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' rlteerjeeat 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 few1 s: i'ftrﬂ. ﬁren+2 :1:
I. fﬁ'UII '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'" — far1.21 
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 altzen divergent by the Comperieon Test. The eese i? = i: — t‘v'e knew that the hnrmenie HIiFiilﬁ 2:1 ,. with f” = Me is divergent. For thie ease
iireﬂhmfnﬂff” = emmmei + life : 1 so this iti 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 pseries with p = 2, is convergent. For this ease
iimmmfﬂlffn : 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 example3 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. NONNEGATIVE 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 NONZEN} 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. NUNNEGATIVE 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 nonnegative 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} nonnegative
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 nonnegative 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. NONINCREASING 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 nonnegative 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}. ...
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 Summer '09
 ganong
 Calculus

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