ALTERNATING SERIES and

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1
Thus far, we have only dealt with series with only positive terms.
An alternating series is one whose terms are alternately positive and negative.
The Alternating series test
∞
The alternating series
∞
∑ ๐‘Ž๐‘› = ∑(−1)๐‘›−1 ๐‘๐‘› = ๐‘1 − ๐‘2 + ๐‘3 − ๐‘4 + โ‹ฏ
๐‘›=1
๐‘๐‘› > 0
๐‘›=1
converges if
๐‘–)
๐‘–๐‘–)
๐‘๐‘›+1 ≤ ๐‘๐‘›
๐‘“๐‘œ๐‘Ÿ ๐‘Ž๐‘™๐‘™ ๐‘› ∈ ๐‘ +
(๐‘‘๐‘’๐‘๐‘Ÿ๐‘’๐‘Ž๐‘ ๐‘–๐‘›๐‘”)
๐‘™๐‘–๐‘š ๐‘๐‘› = 0
๐‘›→∞
Note: The theorem also applies if the first term is negative, since we could simply consider the series
without the first term.
Proof:
2๐‘›๐‘กโ„Ž ๐‘๐‘Ž๐‘Ÿ๐‘ก๐‘–๐‘Ž๐‘™ ๐‘ ๐‘ข๐‘š ๐‘–๐‘ :
โˆŽ
๐‘†2๐‘› = (๐‘1 − ๐‘2 ) + (๐‘3 − ๐‘4 ) + โ‹ฏ + (๐‘2๐‘›−1 − ๐‘2๐‘› )
{๐‘๐‘› } ๐‘–๐‘  ๐‘š๐‘œ๐‘›๐‘œ๐‘ก๐‘œ๐‘›๐‘–๐‘๐‘Ž๐‘™๐‘™๐‘ฆ ๐‘‘๐‘’๐‘๐‘Ÿ๐‘’๐‘Ž๐‘ ๐‘–๐‘›๐‘”, ๐‘2๐‘›−1 − ๐‘2๐‘› ≥ 0 ๐‘“๐‘œ๐‘Ÿ ๐‘Ž๐‘™๐‘™ ๐‘›
⇒ ๐‘†2๐‘› ๐‘–๐‘  ๐‘š๐‘œ๐‘›๐‘œ๐‘ก๐‘œ๐‘›๐‘–๐‘๐‘Ž๐‘™๐‘™๐‘ฆ ๐‘–๐‘›๐‘๐‘Ÿ๐‘’๐‘Ž๐‘ ๐‘–๐‘›๐‘”
(If the first term is negative, consider the series in the absence of the first term.)
โˆŽ
๐‘†2๐‘› = ๐‘1 − (๐‘2 − ๐‘3 ) − (๐‘4 − ๐‘5 ) − โ‹ฏ − (๐‘2๐‘›−2 − ๐‘2๐‘›−1 ) − ๐‘2๐‘›
{๐‘๐‘› } ๐‘–๐‘  ๐‘š๐‘œ๐‘›๐‘œ๐‘ก๐‘œ๐‘›๐‘–๐‘๐‘Ž๐‘™๐‘™๐‘ฆ ๐‘‘๐‘’๐‘๐‘Ÿ๐‘’๐‘Ž๐‘ ๐‘–๐‘›๐‘”, ๐‘2๐‘›−2 − ๐‘2๐‘›−1 ≥ 0 ๐‘“๐‘œ๐‘Ÿ ๐‘Ž๐‘™๐‘™ ๐‘›
⇒ ๐‘†2๐‘› ≤ ๐‘1
๐‘†๐‘’๐‘ž๐‘ข๐‘’๐‘›๐‘๐‘’ {๐‘†2๐‘› } ๐‘–๐‘  ๐‘š๐‘œ๐‘›๐‘œ๐‘ก๐‘œ๐‘›๐‘–๐‘๐‘Ž๐‘™๐‘™๐‘ฆ ๐‘–๐‘›๐‘๐‘Ÿ๐‘’๐‘Ž๐‘ ๐‘–๐‘›๐‘” ๐‘Ž๐‘›๐‘‘ ๐‘๐‘œ๐‘ข๐‘›๐‘‘๐‘’๐‘‘.
๐‘‡โ„Ž๐‘ข๐‘  ๐‘กโ„Ž๐‘’ ๐‘ ๐‘’๐‘ž๐‘ข๐‘’๐‘›๐‘๐‘’ {๐‘†2๐‘› } โ„Ž๐‘Ž๐‘  ๐‘Ž ๐‘™๐‘–๐‘š๐‘–๐‘ก ๐‘๐‘ฆ ๐‘กโ„Ž๐‘’ ๐‘€๐‘œ๐‘›๐‘œ๐‘ก๐‘œ๐‘›๐‘’ ๐ถ๐‘œ๐‘›๐‘ฃ๐‘’๐‘Ÿ๐‘”๐‘’๐‘›๐‘๐‘’ ๐‘‡โ„Ž๐‘’๐‘œ๐‘Ÿ๐‘’๐‘š.
๐ฟ๐‘’๐‘ก ๐‘† = lim ๐‘†2๐‘›
๐‘›→∞
โˆŽ ๐‘๐‘œ๐‘ค ๐‘ค๐‘’ ๐‘๐‘œ๐‘›๐‘ ๐‘–๐‘‘๐‘’๐‘Ÿ ๐‘กโ„Ž๐‘’ ๐‘œ๐‘‘๐‘‘ ๐‘๐‘Ž๐‘Ÿ๐‘ก๐‘–๐‘Ž๐‘™ ๐‘ ๐‘ข๐‘š๐‘ : ๐‘†2๐‘›+1 = ๐‘†2๐‘› + ๐‘2๐‘›+1
lim ๐‘†2๐‘›+1 = lim ๐‘†2๐‘› + lim ๐‘๐‘› = ๐‘†
๐‘›→∞
∴
๐‘›→∞
๐‘›→∞
๐‘๐‘› → 0 ๐‘๐‘ฆ ๐‘œ๐‘ข๐‘Ÿ โ„Ž๐‘ฆ๐‘๐‘œ๐‘กโ„Ž๐‘’๐‘ ๐‘’๐‘ .
๐‘‡โ„Ž๐‘ข๐‘  ๐‘กโ„Ž๐‘’ ๐‘’๐‘ฃ๐‘’๐‘› ๐‘Ž๐‘›๐‘‘ ๐‘œ๐‘‘๐‘‘ ๐‘ ๐‘’๐‘Ÿ๐‘–๐‘’๐‘  ๐‘๐‘œ๐‘กโ„Ž ๐‘๐‘œ๐‘›๐‘ฃ๐‘’๐‘Ÿ๐‘”๐‘’ ๐‘ก๐‘œ ๐‘กโ„Ž๐‘’ ๐‘ ๐‘Ž๐‘š๐‘’ ๐‘™๐‘–๐‘š๐‘–๐‘ก ๐‘–๐‘“ ๐‘Ž๐‘›๐‘‘ ๐‘œ๐‘›๐‘™๐‘ฆ ๐‘–๐‘“ lim ๐‘๐‘› = 0
๐‘›→∞
Q.E.D.
2
๐ธ๐‘ฅ๐‘Ž๐‘š๐‘๐‘™๐‘’:
1 1 1
Show that 1 − + − + โ‹ฏ
2 3 4
∞
(−1)๐‘›−1
=∑
converges.
๐‘›
๐‘›=1
๐‘†๐‘–๐‘›๐‘๐‘’
1
1
< , ๐‘กโ„Ž๐‘’ ๐‘ ๐‘’๐‘Ÿ๐‘–๐‘’๐‘  ๐‘ ๐‘Ž๐‘ก๐‘–๐‘ ๐‘“๐‘–๐‘’๐‘  ๐‘๐‘›+1 ≤ ๐‘๐‘› ๐‘“๐‘œ๐‘Ÿ ๐‘Ž๐‘™๐‘™ ๐‘› ∈ ๐‘ +
๐‘›+1 ๐‘›
๐ด๐‘™๐‘ ๐‘œ ๐‘™๐‘–๐‘š ๐‘๐‘› = ๐‘™๐‘–๐‘š
๐‘›→∞
∞
∴ ∑
๐‘›=1
1
(Alternating series for which ๐‘๐‘› = )
๐‘›
๐‘›→∞
1
=0
๐‘›
(−1)๐‘›−1
๐‘๐‘œ๐‘›๐‘ฃ๐‘’๐‘Ÿ๐‘”๐‘’๐‘  ๐‘๐‘ฆ ๐‘กโ„Ž๐‘’ ๐ด๐‘™๐‘ก๐‘’๐‘Ÿ๐‘›๐‘Ž๐‘ก๐‘–๐‘›๐‘” ๐‘†๐‘’๐‘Ÿ๐‘–๐‘’๐‘  ๐‘‡๐‘’๐‘ ๐‘ก
๐‘›
∞
(๐‘’๐‘ฃ๐‘’๐‘› ๐‘กโ„Ž๐‘œ๐‘ข๐‘”โ„Ž ∑
๐‘›=1
1
๐‘–๐‘  ๐‘›๐‘œ๐‘ก ๐‘๐‘œ๐‘›๐‘ฃ๐‘’๐‘Ÿ๐‘”๐‘’๐‘›๐‘ก)
๐‘›
The Alternating Series Remainder Estimates
A partial sum Sn of any convergent series can be used as an approximation to the total sum S.
The error involved in using ๐‘† ≈ ๐‘†๐‘› , for the sum of a converging alternating series is the remainder
๐‘…๐‘› = ๐‘† − ๐‘†๐‘›
and
|๐‘…๐‘› | = |๐‘† − ๐‘†๐‘› | ≤ ๐‘๐‘›+1
(error = remainder = truncation error)
Proof:
∞
๐‘…๐‘› = ๐‘† − ๐‘†๐‘› = ∑ (−1)๐‘˜−1 ๐‘๐‘˜ = (−1)๐‘› ๐‘๐‘›+1 + (−1)๐‘›+1 ๐‘๐‘›+2 + โ‹ฏ
๐‘˜=๐‘›+1
โˆŽ
๐‘…๐‘› = (−1)๐‘› [(๐‘๐‘›+1 − ๐‘๐‘›+2 ) + (๐‘๐‘›+3 − ๐‘๐‘›+4 ) + โ‹ฏ ]
{๐‘๐‘› } ๐‘–๐‘  ๐‘š๐‘œ๐‘›๐‘œ๐‘ก๐‘œ๐‘›๐‘–๐‘๐‘Ž๐‘™๐‘™๐‘ฆ ๐‘‘๐‘’๐‘๐‘Ÿ๐‘’๐‘Ž๐‘ ๐‘–๐‘›๐‘”, ๐‘๐‘›+๐‘Ÿ − ๐‘๐‘›+๐‘Ÿ+1 ≥ 0 ๐‘“๐‘œ๐‘Ÿ ๐‘Ž๐‘™๐‘™ ๐‘›
⇒ |๐‘…๐‘› | = (๐‘๐‘›+1 − ๐‘๐‘›+2) + (๐‘๐‘›+3 − ๐‘๐‘›+4 ) + โ‹ฏ ≥ 0
โˆŽ
๐‘…๐‘› = (−1)๐‘› {๐‘๐‘›+1 − [(๐‘๐‘›+2 − ๐‘๐‘›+3 ) + (๐‘๐‘›+4 − ๐‘๐‘›+5 ) + โ‹ฏ ]}
≥ 0 ๐‘“๐‘œ๐‘Ÿ ๐‘Ž๐‘™๐‘™ ๐‘› [{๐‘๐‘› } ๐‘–๐‘  ๐‘š๐‘œ๐‘›๐‘œ๐‘ก๐‘œ๐‘›๐‘–๐‘๐‘Ž๐‘™๐‘™๐‘ฆ ๐‘‘๐‘’๐‘๐‘Ÿ๐‘’๐‘Ž๐‘ ๐‘–๐‘›๐‘”]
⇒ |๐‘…๐‘› | ≤ ๐‘๐‘›+1
Q.E.D.
3
∞
EX: Find the sum of ∑
๐‘›=1
0<
(−1)๐‘›−1
correct to 3 decimal places.
๐‘›!
1
1
<
(๐‘› + 1)! ๐‘›!
๐ด๐‘™๐‘ ๐‘œ,
0<
๐‘†๐‘–๐‘›๐‘๐‘’ lim
๐‘›→∞
( ๐‘๐‘› =
1
)
๐‘›!
∴ ๐‘๐‘›+1 ≤ ๐‘๐‘› ๐‘“๐‘œ๐‘Ÿ ๐‘Ž๐‘™๐‘™ ๐‘› ∈ ๐‘ +
1
1
<
๐‘›!
๐‘›
1
=0 ⇒
๐‘›
lim
๐‘›→∞
1
=0
๐‘›!
๐‘๐‘ฆ ๐‘กโ„Ž๐‘’ ๐‘†๐‘ž๐‘ข๐‘’๐‘’๐‘ง๐‘’ ๐‘‡โ„Ž๐‘’๐‘œ๐‘Ÿ๐‘’๐‘š
∴ ๐‘กโ„Ž๐‘’ ๐‘ ๐‘’๐‘Ÿ๐‘–๐‘’๐‘  ๐‘๐‘œ๐‘›๐‘ฃ๐‘’๐‘Ÿ๐‘”๐‘’๐‘  ๐‘๐‘ฆ ๐‘กโ„Ž๐‘’ ๐ด๐‘™๐‘ก๐‘’๐‘Ÿ๐‘›๐‘Ž๐‘ก๐‘–๐‘›๐‘” ๐‘†๐‘’๐‘Ÿ๐‘–๐‘’๐‘  ๐‘‡๐‘’๐‘ ๐‘ก
๐‘† =1−
1 1 1
1
1
1
+ −
+
−
+
+โ‹ฏ
2 6 24 120 720 5040
๐‘7 =
1
1
<
5040
2000
= 0.0005 ๐‘š๐‘Ž๐‘›๐‘ข๐‘Ž๐‘™๐‘™๐‘ฆ
๐‘†6 = 1 −
1 1 1
1
1
+ −
+
−
= 0.631944
2 6 24 120 720
๐ต๐‘ฆ ๐ธ๐‘ ๐‘ก๐‘–๐‘š๐‘Ž๐‘ก๐‘–๐‘œ๐‘› ๐‘‡โ„Ž๐‘’๐‘œ๐‘Ÿ๐‘’๐‘š, |๐‘† − ๐‘†6 | ≤ ๐‘7
∴ 0.631944 −
1
1
≤ ๐‘† ≤ 0.631944 +
⇒ 0.6317456 ≤ ๐‘† ≤ 0.6321424
5040
5040
๐‘† ≈ ๐‘†6 = 0. 632
(3 ๐‘‘๐‘’๐‘๐‘–๐‘š๐‘Ž๐‘™ ๐‘๐‘™๐‘Ž๐‘๐‘’๐‘ )
∞
EX: Aproximate the sum of ∑
๐‘›=1
(−1)๐‘›
by its first 5 terms (5th partial sum).
๐‘›2
๐น๐‘–๐‘›๐‘‘ ๐‘กโ„Ž๐‘’ ๐‘–๐‘›๐‘ก๐‘’๐‘Ÿ๐‘ฃ๐‘Ž๐‘™ ๐‘–๐‘› ๐‘คโ„Ž๐‘–๐‘โ„Ž ๐‘ ๐‘ข๐‘š ๐‘† ๐‘™๐‘–๐‘’๐‘ .
๐‘†5 = −1 +
1 1 1
1
− +
−
≈ −0.838611
4 9 16 25
1
1
≤
(๐‘› + 1)2 ๐‘›2
⇒
lim
๐‘›→∞
1
=0
๐‘›2
∴ ๐‘๐‘›+1 ≤ ๐‘๐‘› ๐‘“๐‘œ๐‘Ÿ ๐‘Ž๐‘™๐‘™ ๐‘› ∈ ๐‘ +
∴ ๐‘กโ„Ž๐‘’ ๐‘ ๐‘’๐‘Ÿ๐‘–๐‘’๐‘  ๐‘๐‘œ๐‘›๐‘ฃ๐‘’๐‘Ÿ๐‘”๐‘’๐‘  ๐‘๐‘ฆ ๐‘กโ„Ž๐‘’ ๐ด๐‘™๐‘ก๐‘’๐‘Ÿ๐‘›๐‘Ž๐‘ก๐‘–๐‘›๐‘” ๐‘†๐‘’๐‘Ÿ๐‘–๐‘’๐‘  ๐‘‡๐‘’๐‘ ๐‘ก
๐ธ๐‘Ÿ๐‘Ÿ๐‘œ๐‘Ÿ ๐‘๐‘œ๐‘ข๐‘›๐‘‘ ๐‘–๐‘  ๐‘”๐‘–๐‘ฃ๐‘’๐‘› ๐‘๐‘ฆ: |๐‘† − ๐‘†5 | = |๐‘…5 | ≤
๐‘†๐‘œ ๐‘Ž๐‘๐‘ก๐‘ข๐‘Ž๐‘™ ๐‘ ๐‘ข๐‘š ๐‘™๐‘–๐‘’๐‘  ๐‘๐‘’๐‘ก๐‘ค๐‘’๐‘’๐‘›
1
≈ 0.027778
36
− 0.838611 − 0.027778 ≤ ๐‘† ≤ −0.838611 + 0.027778
4
ABSOLUTE AND CONDITIONAL CONVERGENCE (for loco series +, - ,+,+,+,-,-,-)
∞
Given ๐€๐๐˜ series
∞
∑ ๐‘Ž๐‘› we can consider the corresponding series ∑|๐‘Ž๐‘› | = |๐‘Ž1 | + |๐‘Ž2 | + โ‹ฏ
๐‘›=1
๐‘›=1
Absolute Convergence
∞
∞
The series ∑ ๐‘Ž๐‘› is ๐š๐›๐ฌ๐จ๐ฅ๐ฎ๐ญ๐ฅ๐ฒ ๐œ๐จ๐ง๐ฏ๐ž๐ซ๐ ๐ž๐ง๐ญ if ∑|๐‘Ž๐‘› | converges.
๐‘›=1
๐‘›=1
If ๐‘Ž๐‘› ≥ 0 for all n, absolute convergence is the same as convergence.
Conditional Convergence
∞
∞
∞
The series ∑ ๐‘Ž๐‘› is ๐œ๐จ๐ง๐๐ข๐ญ๐ข๐จ๐ง๐š๐ฅ๐ฅ๐ฒ ๐œ๐จ๐ง๐ฏ๐ž๐ซ๐ ๐ž๐ง๐ญ if ∑ ๐‘Ž๐‘› converges but ∑|๐‘Ž๐‘› | diverges .
๐‘›=1
๐‘›=1
๐‘›=1
∞
(−1)๐‘›−1
๐ธ๐‘‹: ∑
๐‘–๐‘  ๐‘๐‘œ๐‘›๐‘‘๐‘–๐‘ก๐‘–๐‘Ž๐‘™๐‘™๐‘ฆ ๐‘๐‘œ๐‘›๐‘ฃ๐‘’๐‘Ÿ๐‘”๐‘’๐‘›๐‘ก
๐‘›
๐‘›=1
Theorem of Absolute Convergence
∞
If ∑ ๐‘Ž๐‘› is absolutely convergent then it is also convergent.
๐‘›=1
Proof:
by definition of absolute value: − |๐‘Ž๐‘› | ≤ ๐‘Ž๐‘› ≤ |๐‘Ž๐‘› | ⇒ 0 ≤ ๐‘Ž๐‘› + |๐‘Ž๐‘› | ≤ 2|๐‘Ž๐‘› |
∞
∞
If ∑ ๐‘Ž๐‘› is absolutely convergent then 2 ∑|๐‘Ž๐‘› | is convergent.
๐‘›=1
๐‘›=1
∴ by the Comparison Test ∑∞
๐‘›=1(๐‘Ž๐‘› + |๐‘Ž๐‘› |) is also convergent.
∞
๐‘†๐‘–๐‘›๐‘๐‘’ ∑ ๐‘Ž๐‘› ๐‘–๐‘  ๐‘กโ„Ž๐‘’ ๐‘‘๐‘–๐‘“๐‘“๐‘’๐‘Ÿ๐‘’๐‘›๐‘๐‘’ ๐‘œ๐‘“ ๐‘ก๐‘ค๐‘œ ๐‘๐‘œ๐‘›๐‘ฃ๐‘’๐‘Ÿ๐‘”๐‘’๐‘›๐‘ก ๐‘ ๐‘’๐‘Ÿ๐‘–๐‘’๐‘ 
๐‘›=1
∞
∞
∞
∑ ๐‘Ž๐‘› = ∑(๐‘Ž๐‘› + |๐‘Ž๐‘› |) − ∑|๐‘Ž๐‘› |
๐‘›=1
๐‘›=1
๐‘›=1
∞
∴ ∑ ๐‘Ž๐‘› is convergent
๐‘›=1
Q.E.D.
Absolute convergence is a “stronger” type of convergence. Series that are absolutely convergent are
guaranteed to be convergent. However, series that are convergent may or may not be absolutely
convergent (they do converge, but the sum is loco).
5
∞
EX: Show that ∑
๐‘›=1
∞
∑
๐‘›=1
cos ๐‘›
is convergent.
๐‘›2
cos ๐‘›
cos 1 cos 2
=
+ 2 +โ‹ฏ
2
๐‘›
12
2
๐‘ก๐‘’๐‘Ÿ๐‘š๐‘  ๐‘ค๐‘–๐‘กโ„Ž ๐‘‘๐‘–๐‘“๐‘“๐‘’๐‘Ÿ๐‘’๐‘›๐‘ก ๐‘ ๐‘–๐‘”๐‘›๐‘ , ( ๐‘›๐‘œ๐‘ก ๐‘Ž๐‘™๐‘ก๐‘’๐‘Ÿ๐‘›๐‘Ž๐‘ก๐‘–๐‘›๐‘” ๐‘ ๐‘’๐‘Ÿ๐‘–๐‘’๐‘ ).
∞
cos ๐‘›
1
๐ป๐‘œ๐‘ค๐‘’๐‘ฃ๐‘’๐‘Ÿ | 2 | ≤ 2
๐‘›
๐‘›
๐‘“๐‘œ๐‘Ÿ ๐‘Ž๐‘™๐‘™ ๐‘› ∈ ๐‘
+
๐‘Ž๐‘›๐‘‘ ∑
๐‘›=1
∞
∴ ๐‘๐‘ฆ ๐‘กโ„Ž๐‘’ ๐ถ๐‘œ๐‘š๐‘๐‘Ž๐‘Ÿ๐‘–๐‘ ๐‘œ๐‘› ๐‘‡๐‘’๐‘ ๐‘ก,
∑|
๐‘›=1
1
๐‘–๐‘  ๐‘๐‘œ๐‘›๐‘ฃ๐‘’๐‘Ÿ๐‘”๐‘’๐‘›๐‘ก.
๐‘›2
cos ๐‘›
| ๐‘–๐‘  ๐‘๐‘œ๐‘›๐‘ฃ๐‘’๐‘Ÿ๐‘”๐‘’๐‘›๐‘ก,
๐‘›2
๐‘Ž๐‘›๐‘‘ ๐‘๐‘ฆ ๐‘กโ„Ž๐‘’ ๐‘‡โ„Ž๐‘’๐‘œ๐‘Ÿ๐‘’๐‘š ๐‘œ๐‘“ ๐ด๐‘๐‘ ๐‘œ๐‘™๐‘ข๐‘ก๐‘’ ๐ถ๐‘œ๐‘›๐‘ฃ๐‘’๐‘Ÿ๐‘”๐‘’๐‘›๐‘๐‘’,
∞
๐‘ ๐‘œ ๐‘–๐‘ 
∑
๐‘›=1
cos ๐‘›
๐‘›2
So what is important about absolute and conditional convergence?
Algebra: a + b = b + a. If we have a finite sum ∑ ๐‘Ž๐‘› , we can also reorder the terms without affecting the
sum. Infinite series which are absolute convergent behave like finite series, so for these we can again
reorder the terms of the series without affecting the sum. However, the same is not true for conditionally
convergent series!
๐ธ๐‘‹:
1 1 1 1 1
๐‘†=1− + − + − +โ‹ฏ
2 3 4 5 6
⇒
1
1 1 1 1
๐‘† = − + − +โ‹ฏ
2
2 4 6 8
๐‘Ž๐‘‘๐‘‘ ๐‘กโ„Ž๐‘’๐‘š ๐‘ก๐‘œ๐‘”๐‘’๐‘กโ„Ž๐‘’๐‘Ÿ
3
1 1 1
2
1 1 1 1 1
๐‘† = 1 + 0 + − + + 0 โ‹ฏ ⇒ ๐‘† = (1 − + − + − + โ‹ฏ )
2
3 2 5
3
2 3 4 5 6
Thus we get a rearrangement of the original series with a different sum!
โ— If ∑ ๐‘Ž๐‘› is absolutely convergent and its value is s then any rearrangement of ∑ ๐‘Ž๐‘› will also have
a value of s.
โ— If ∑ ๐‘Ž๐‘› is conditionally convergent and r is any real number then there is a rearrangement of
∑ ๐‘Ž๐‘› whose value will be r. NO KIDDING !!!!!! But the series is convergent !!!!!!!!!! Yet, we don’t
know the sum. (proof shown by Riemann) -๏Œ
(−1)๐‘›−1
โ–ช We found that alternating harmonic series ∑
is conditionally convergent, so no matter what value
๐‘›
we choose there is some rearrangement of terms that will give that value. Unbelievable, right. But true.
Note that above fact does not tell us what that rearrangement must be, ONLY that it does exist.
(−1)๐‘›
โ–ช We found that alternating series ∑ 2 is absolutely convergent and so no matter how we rearrange the
๐‘›
terms of this series we’ll always get the same value. -๏Š . In fact it can be shown that the value of that series
is
∞
∑
๐‘›=1
(−1)๐‘›
๐œ‹2
=
−
๐‘›2
12
6
ALTERNATING and LOCO SERIES PROBLEMS
1. Determine if each of the following series are absolute convergent, conditionally convergent or divergent.
∞
(−1)๐‘›−1
(๐‘Ž) ∑
๐‘›
๐‘›=1
∞
(−1)๐‘›
( ๐‘) ∑
๐‘›2
๐‘›=1
∞
sin ๐‘›
(๐‘) ∑ 3
๐‘›
๐‘›=1
∞
(−1)๐‘› ๐‘›2
(๐‘‘) ∑ 2
๐‘› +5
๐‘›=1
∞
(−1)๐‘›−3 √๐‘›
๐‘’) ∑
๐‘›+4
๐‘›=1
∞
๐‘“) ∑
๐‘›=2
cos(๐‘›๐œ‹)
√๐‘›
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