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Exercice 14 Etudier la nature des séries de terme général et calculer leur somme : 1 ( ) 2

  • Comment calculer ? ?

    ? [terme général d'une suite arithmétique] = [nombre de termes] × [premier terme] + [dernier terme] 2 .

  • Comment calculer la somme Sigma ?

    Somme simple
    Le symbole ? (sigma) s'utilise pour désigner de manière générale la somme de plusieurs termes. Ce symbole est généralement accompagné d'un indice que l'on fait varier de façon à englober tous les termes qui doivent être considérés dans la somme.

  • Comment calculer la somme de K ?

    k = n (n + 1) 2 . La variable k est appelée indice de la somme; on utilise aussi fréquemment la lettre i comme variable d'indice.
  • Deux séries sont dites de même nature lorsqu'elles sont toutes les deux convergentes ou toutes les deux divergentes. Déterminer la nature d'une série c'est déterminer si elle converge ou si elle diverge. vn converge.

Lecture 24Section 11.4 Absolute and Conditional

Convergence; Alternating Series

Jiwen He

1 Convergence Tests

Basic Series that Converge or Diverge

Basic Series that Converge

Geometric series:

?xk,if|x|<1 p-series:?1k p,ifp >1

Basic Series that Diverge

Any series

?a kfor which limk→∞ak?= 0 p-series:?1k

Convergence Tests (1)

Basic Test for Convergence

Keep in Mind that, ifak?0, then the series?akdiverges; therefore there is no reason to apply any special convergence test.Examples1.? xkwith|x| ≥1 (e.g,?(-1)k)divergesincexk?0. [1ex] ?kk+ 1divergessincekk+1→1?= 0. [1ex]?? 1-1k k divergessince a k=?1-1k k→e-1?= 0.

Convergence Tests (2)

Comparison Tests

Rational termsare most easily handled bybasic comparisonorlimit comparison withp-series?1/kp

Basic Comparison Test1

12k3+ 1converges by comparison with?1k

3? k3k

5+ 4k4+ 7converges

by comparison with ?1k 2? 1k

3-k2converges by comparison with?2k

3?13k+ 1diverges by comparison with?13(k+ 1)?

1ln(k+ 6)diverges by

comparison with ?1k+ 6

Limit Comparison Test

?1k

3-1converges by comparison with?1k

3.?3k2+ 2k+ 1k

3+ 1diverges

by comparison with ?3k

5⎷k+ 1002k2⎷k-9⎷k

converges by comparison with 52k2

Convergence Tests (3)

Root Test and Ratio Test

Theroot testis used only ifpowersare involved.

Root Test

?k22 kconverges: (ak)1/k=12

·?k1/k?2→12

·1?1(lnk)kconverges: (ak)1/k=

1lnk→0??

1-1k k2 converges: (ak)1/k=?

1 +(-1)k

k→e-1

Convergence Tests (4)

Root Test and Ratio Test

Theratio testis effective withfactorialsand with combinations of powers and factorials.

Ratio Comparison Test?k22

kconverges:ak+1a k=12

·(k+1)2k

2→12

1k!converges:ak+1a

k=1k+1→0 k10 kconverges:ak+1a k=110

·k+1k

→110 kkk!diverges:ak+1a k=?1 +1k k→e 2k3 k-2kconverges:ak+1a k= 2·1-(2/3)k3-2(2/3)k→2·13

1⎷k!converges:ak+1a

k= ?1 k+1→0

2 Absolute Convergence

2.1 Absolute Convergence

Absolute Convergence2

Absolute Convergence

A series?akis said toconverge absolutelyif?|ak|converges. if ?|ak|converges, then?akconverges. i.e., absolutely convergent series are convergent.

Alternatingp-Series withp >1?(-1)kk

p,p >1,converge absolutelybecause?1k pconverges.? k=1(-1)k+1k

2= 1-12

2+13 2-14

2- ···converge absolutely.

Geometric Series with-1< x <1?(-1)j(k)xk,-1< x <1,converge absolutelybecause?|x|kconverges. ?1-12 -12 2+12 3-12 4+12 5+12

6- ···converge absolutely.

Conditional Convergence

Conditional Convergence

A series?akis said toconverge conditionallyif?akconverges while?|ak| diverges. pdiverges.? k=1(-1)k+1k = 1-12 +13 -14 - ···converge conditionally.

3 Alternating Series

Alternating Series

Alternating Series

Let{ak}be a sequence ofpositivenumbers.?(-1)kak=a0-a1+a2-a3+a4- ··· is called analternating series.

Alternating Series Test

Let{ak}be adecreasingsequence ofpositivenumbers.

Ifak→0,then?(-1)kakconverges.Alternatingp-Series withp >0?(-1)kk p,p >0,convergesincef(x) =1x pisdecreasing, i.e.,f?(x) =-px p+1>

0 for?x >0, and limx→∞f(x) = 0.?∞?

k=1(-1)k+1k = 1-12 +13 -14 convergeconditionally.3

Examples

(-1)k2k+ 1,convergesincef(x) =12x+ 1isdecreasing, i.e.,f?(x) =-2(2x+ 1)2>

0 for?x >0, and limx→∞f(x) = 0.

(-1)kkk

2+ 10,convergesincef(x) =xx

2+ 10isdecreasing, i.e.,f?(x) =-x2-10(x2+ 10)2>

0, for?x >⎷10, and lim

x→∞f(x) = 0. An Estimate for Alternating SeriesAn Estimate for Alternating Series Let{ak}be adecreasingsequence ofpositivenumbers that tends to 0 and let

L=∞?

k=0(-1)kak. Then the sumLlies between consecutivepartial sumssn, s n+1,sn< L < sn+1,ifnis odd;sn+1< L < sn,ifnis even. and thussnapproximatesLto withinan+1 |L-sn|< an+1.

Example

Findsnto approximate∞?

k=1(-1)k+1k = 1-12 +13

···within 10-2.

Set k=1(-1)k+1k k=0(-1)kk+ 1. For|L-sn|<10-2, we want a n+1=1(n+ 1) + 1<10-2?n+ 2>102?n >98.

Thenn= 99 and the 99th partial sums100is

s

99= 1-12

+13 -14 +···+199 -1100 ≈0.6882.

From the estimate

|L-s99|< a100=1101 ≈0.00991. we conclude that s

99≈0.6882<∞?

k=1(-1)k+1k = ln2<0.6981≈s1004

Example

Findsnto approximate∞?

k=0(-1)k+1(2k+ 1)!= 1-13! +15!

···within 10-2.

For|L-sn|<10-2, we want

a n+1=1(2(n+ 1) + 1)!<10-2?n≥1.

Thenn= 1 and the 2nd partial sums2is

s

1= 1-13!

≈0.8333

From the estimate|L-s1|< a2=15!

≈0.0083. we conclude that s

1≈0.8333<∞?

k=0(-1)k+1(2k+ 1)!= sin1<0.8416≈s2

4 Rearrangements

Why Absolute Convergence Matters: Rearrangements (1)

Rearrangement of Absolute Convergence Series

k=0(-1)k2 k= 1-12 +12 2-12 3+12 4-12

5+···=23

absolutely

Rearrangement1 +12

2-12 +12 4+12 6-12 3+12 8+12 10-12

5···? = =23

Theorem 2.All rearrangements of an absolutely convergent series converge absolutely to the same sum. Why Absolute Convergence Matters: Rearrangements (2)

Rearrangement of Conditional Convergence Series

k=1(-1)k+1k = 1-12 +13 -14 +15 -16 +···= ln2conditionally

Rearrangement1 +13

-12 +15 +17 -14 +19 +111
-16

···? =?= ln2

Multiply the original series by

12 12 k=1(-1)k+1k =12 -14 +16 -18 +110
+···=12 ln2

Adding the two series, we get the rearrangement

k=1(-1)k+1k +12 k=1(-1)k+1k = 1 +13 -12 +15 +17 -14 +···=32 ln2

Remark5

•A series that is onlyconditionallyconvergent can be rearranged to converge

toany numberwe please.•It can also be arranged todivergeto +∞or-∞, or even to oscillate

between any two bounds we choose.

Outline

Contents

1 Convergence Tests 1

2 Absolute Convergence 2

2.1 Absolute Convergence . . . . . . . . . . . . . . . . . . . . . . . .2

3 Alternating Series 3

4 Rearrangements 56

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