Unformatted text preview: INFINITE SERIES — SUMMARY ' GEOMETRJO seams: WITH RATIO 1' (AND c are 0) m 1 N _ V+1
Zcr" = c(1+r+1'2i1"r’+1"+...)={cfmverges T< Since 2"" 5 SN :1 l“
n==0 diverges M21 “=0 l—r
psnnms
i0)” _ i}— _ 1+(1)’+(1)P+<1)P+ _ {converge p>1 ShowthisviaIntegraiTeet.
n "an? __ 2 3 4 _ diverges pgl pr=1.it’scalledtheharmonicseries. n=l nthTERM TEST FOR DWERGENCE The Test: If 11m...“ an 9'5 0 or lino...” au DNE, then 2 a... diverges
Because: If E a,I converges. then 11111,...” a... = 0
Warning: If limﬂnoo aﬂ = 0, then it is possible that Z a" converges and it is possible that Z a" diverges Remark: The n"I can show divergence but can NOT show convergence.
2 a... is absolutely convergent 4: [ z ]a... converges ]
Z on is conditionally convergent d: [ Z {anl diverges AND z a" converges ]
2a" isdiLergelt <= [zen diverges]
Theorem: If Z Ianl converges then 2a,. converges.
So we get for free: If E on diverges, then 2 la...) diverges MU'I‘UALLY EXCLUSIVE AND EXHAUSTIVE POSSIBILITIES 2 an is absolutely convergent 4:. [ 2 Ian] converga “g” 2 an converga ]
2 an is conditionally convergent 4: [ 2 [an[ diverga AND 2 a... converges ]
E aYl is divergent => [ 2 a“ diverges implies 2 1a,.) diverge ] PROBLEM: we need to ﬁgure out if an inﬁnite series 2 a... is: absolutely convergent. conditionally convergent, or divergent.
SOLUTION: we apply one of the below TESTS that will give us the answer. Which one . . . well, pattern recognition time. Sometimes
more than one test will work! For some of the tests. we need to ﬁnd the appropriate 2 b," which is usually a. welllmown series (like
a. geometric series or pseries) that we know whether it converges or diverges. NAME STATEMENT OF TEST POSITIVETERMED SERIES TESTS
Ea... where (1‘20 VneN Key Idea 2a,. converge {=9 {5N})°v°=1 is bounded above (since a, 2 0 4: 3,, /' ) Integral Test Let f: [1. co) —’ R be continuous. positive, and nonincreasing function with f (n) = anVn E N.
Then [2a.u converge e: f1” f (z)d:r. converges] . Comparison’I‘est [DgansannZIVochbnconw ]== [Zanconm] (CT) [0Sbn<anV172No&zbndivg. ]=>[Za.ndivg.] Limit Comparison Test Let b, > 0 and lini.,._.m at = L. (LCT) If 0 < L < co, then [ Ea." conv. 4:» 2b,, conv. ] (you DO need to memorize this one)
If L = 0. then [ 2 b" conv. = 2a,. conv. ] (you do not have to memorize this one)
If L = 00. then [ 2 b“ divg. => 2 an divg. ] (you do not have to memorize this one) ARBITRARY—TERMED SERIES TESTS
2a,. where —oo<n,,<oo VneN Ratio Test Letp = umﬂnml‘gﬁ
. . l
Rectum Letp = 11mm» v [all "°=‘° 1mm tall». 0 S p < 1 => 2 an converges absolutely
p = 1 => test is inconclusive
1 < p S 00 = z a... diverges (by n‘“ term test for divergence) ALTERNATING SERIES TEST
2'1" = :(1)"Um where u" > 0 V7: EN, mother words an = (—1)"u" and u,l > 0 Alternating Series Test (AST) [u" > in.“ V7: E N 8; lin1,._.¢,‘,u.n = 0] = [Zan = Z(—l)"u,, conv.] ...
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 Fall '11
 KUSTIN
 Calculus

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