11FSeriesSumAMS

# 11FSeriesSumAMS - INFINITE SERIES — SUMMARY GEOMETRJO...

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Unformatted text preview: INFINITE SERIES — SUMMARY ' GEOMETRJO seams: WITH RATIO 1' (AND c are 0) m 1 N _ V+1 Zcr" = c(1+r-+-1'2-i-1"r’+1"+...)={cfmverges |T|< Since 2"" 5 SN :1 l“ n==0 diverges M21 “=0 l—r p-snnms 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 nth-TERM 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. well-lmown series (like a. geometric series or p-series) that we know whether it converges or diverges. NAME STATEMENT OF TEST POSITIVE-TERMED 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<anV17-2No&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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