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
∞
∑ 𝑎𝑛 = ∑(−1)𝑛−1 𝑏𝑛 = 𝑏1 − 𝑏2 + 𝑏3 − 𝑏4 + ⋯
𝑛=1
𝑏𝑛 > 0
𝑛=1
converges if
𝑖)
𝑖𝑖)
𝑏𝑛+1 ≤ 𝑏𝑛
𝑓𝑜𝑟 𝑎𝑙𝑙 𝑛 ∈ 𝑍 +
(𝑑𝑒𝑐𝑟𝑒𝑎𝑠𝑖𝑛𝑔)
𝑙𝑖𝑚 𝑏𝑛 = 0
𝑛→∞
Note: The theorem also applies if the first term is negative, since we could simply consider the series
without the first term.
Proof:
2𝑛𝑡ℎ 𝑝𝑎𝑟𝑡𝑖𝑎𝑙 𝑠𝑢𝑚 𝑖𝑠:
∎
𝑆2𝑛 = (𝑏1 − 𝑏2 ) + (𝑏3 − 𝑏4 ) + ⋯ + (𝑏2𝑛−1 − 𝑏2𝑛 )
{𝑏𝑛 } 𝑖𝑠 𝑚𝑜𝑛𝑜𝑡𝑜𝑛𝑖𝑐𝑎𝑙𝑙𝑦 𝑑𝑒𝑐𝑟𝑒𝑎𝑠𝑖𝑛𝑔, 𝑏2𝑛−1 − 𝑏2𝑛 ≥ 0 𝑓𝑜𝑟 𝑎𝑙𝑙 𝑛
⇒ 𝑆2𝑛 𝑖𝑠 𝑚𝑜𝑛𝑜𝑡𝑜𝑛𝑖𝑐𝑎𝑙𝑙𝑦 𝑖𝑛𝑐𝑟𝑒𝑎𝑠𝑖𝑛𝑔
(If the first term is negative, consider the series in the absence of the first term.)
∎
𝑆2𝑛 = 𝑏1 − (𝑏2 − 𝑏3 ) − (𝑏4 − 𝑏5 ) − ⋯ − (𝑏2𝑛−2 − 𝑏2𝑛−1 ) − 𝑏2𝑛
{𝑏𝑛 } 𝑖𝑠 𝑚𝑜𝑛𝑜𝑡𝑜𝑛𝑖𝑐𝑎𝑙𝑙𝑦 𝑑𝑒𝑐𝑟𝑒𝑎𝑠𝑖𝑛𝑔, 𝑏2𝑛−2 − 𝑏2𝑛−1 ≥ 0 𝑓𝑜𝑟 𝑎𝑙𝑙 𝑛
⇒ 𝑆2𝑛 ≤ 𝑏1
𝑆𝑒𝑞𝑢𝑒𝑛𝑐𝑒 {𝑆2𝑛 } 𝑖𝑠 𝑚𝑜𝑛𝑜𝑡𝑜𝑛𝑖𝑐𝑎𝑙𝑙𝑦 𝑖𝑛𝑐𝑟𝑒𝑎𝑠𝑖𝑛𝑔 𝑎𝑛𝑑 𝑏𝑜𝑢𝑛𝑑𝑒𝑑.
𝑇ℎ𝑢𝑠 𝑡ℎ𝑒 𝑠𝑒𝑞𝑢𝑒𝑛𝑐𝑒 {𝑆2𝑛 } ℎ𝑎𝑠 𝑎 𝑙𝑖𝑚𝑖𝑡 𝑏𝑦 𝑡ℎ𝑒 𝑀𝑜𝑛𝑜𝑡𝑜𝑛𝑒 𝐶𝑜𝑛𝑣𝑒𝑟𝑔𝑒𝑛𝑐𝑒 𝑇ℎ𝑒𝑜𝑟𝑒𝑚.
𝐿𝑒𝑡 𝑆 = lim 𝑆2𝑛
𝑛→∞
∎ 𝑁𝑜𝑤 𝑤𝑒 𝑐𝑜𝑛𝑠𝑖𝑑𝑒𝑟 𝑡ℎ𝑒 𝑜𝑑𝑑 𝑝𝑎𝑟𝑡𝑖𝑎𝑙 𝑠𝑢𝑚𝑠: 𝑆2𝑛+1 = 𝑆2𝑛 + 𝑏2𝑛+1
lim 𝑆2𝑛+1 = lim 𝑆2𝑛 + lim 𝑏𝑛 = 𝑆
𝑛→∞
∴
𝑛→∞
𝑛→∞
𝑏𝑛 → 0 𝑏𝑦 𝑜𝑢𝑟 ℎ𝑦𝑝𝑜𝑡ℎ𝑒𝑠𝑒𝑠.
𝑇ℎ𝑢𝑠 𝑡ℎ𝑒 𝑒𝑣𝑒𝑛 𝑎𝑛𝑑 𝑜𝑑𝑑 𝑠𝑒𝑟𝑖𝑒𝑠 𝑏𝑜𝑡ℎ 𝑐𝑜𝑛𝑣𝑒𝑟𝑔𝑒 𝑡𝑜 𝑡ℎ𝑒 𝑠𝑎𝑚𝑒 𝑙𝑖𝑚𝑖𝑡 𝑖𝑓 𝑎𝑛𝑑 𝑜𝑛𝑙𝑦 𝑖𝑓 lim 𝑏𝑛 = 0
𝑛→∞
Q.E.D.
2
𝐸𝑥𝑎𝑚𝑝𝑙𝑒:
1 1 1
Show that 1 − + − + ⋯
2 3 4
∞
(−1)𝑛−1
=∑
converges.
𝑛
𝑛=1
𝑆𝑖𝑛𝑐𝑒
1
1
< , 𝑡ℎ𝑒 𝑠𝑒𝑟𝑖𝑒𝑠 𝑠𝑎𝑡𝑖𝑠𝑓𝑖𝑒𝑠 𝑏𝑛+1 ≤ 𝑏𝑛 𝑓𝑜𝑟 𝑎𝑙𝑙 𝑛 ∈ 𝑍 +
𝑛+1 𝑛
𝐴𝑙𝑠𝑜 𝑙𝑖𝑚 𝑏𝑛 = 𝑙𝑖𝑚
𝑛→∞
∞
∴ ∑
𝑛=1
1
(Alternating series for which 𝑏𝑛 = )
𝑛
𝑛→∞
1
=0
𝑛
(−1)𝑛−1
𝑐𝑜𝑛𝑣𝑒𝑟𝑔𝑒𝑠 𝑏𝑦 𝑡ℎ𝑒 𝐴𝑙𝑡𝑒𝑟𝑛𝑎𝑡𝑖𝑛𝑔 𝑆𝑒𝑟𝑖𝑒𝑠 𝑇𝑒𝑠𝑡
𝑛
∞
(𝑒𝑣𝑒𝑛 𝑡ℎ𝑜𝑢𝑔ℎ ∑
𝑛=1
1
𝑖𝑠 𝑛𝑜𝑡 𝑐𝑜𝑛𝑣𝑒𝑟𝑔𝑒𝑛𝑡)
𝑛
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 𝑆 ≈ 𝑆𝑛 , for the sum of a converging alternating series is the remainder
𝑅𝑛 = 𝑆 − 𝑆𝑛
and
|𝑅𝑛 | = |𝑆 − 𝑆𝑛 | ≤ 𝑏𝑛+1
(error = remainder = truncation error)
Proof:
∞
𝑅𝑛 = 𝑆 − 𝑆𝑛 = ∑ (−1)𝑘−1 𝑏𝑘 = (−1)𝑛 𝑏𝑛+1 + (−1)𝑛+1 𝑏𝑛+2 + ⋯
𝑘=𝑛+1
∎
𝑅𝑛 = (−1)𝑛 [(𝑏𝑛+1 − 𝑏𝑛+2 ) + (𝑏𝑛+3 − 𝑏𝑛+4 ) + ⋯ ]
{𝑏𝑛 } 𝑖𝑠 𝑚𝑜𝑛𝑜𝑡𝑜𝑛𝑖𝑐𝑎𝑙𝑙𝑦 𝑑𝑒𝑐𝑟𝑒𝑎𝑠𝑖𝑛𝑔, 𝑏𝑛+𝑟 − 𝑏𝑛+𝑟+1 ≥ 0 𝑓𝑜𝑟 𝑎𝑙𝑙 𝑛
⇒ |𝑅𝑛 | = (𝑏𝑛+1 − 𝑏𝑛+2) + (𝑏𝑛+3 − 𝑏𝑛+4 ) + ⋯ ≥ 0
∎
𝑅𝑛 = (−1)𝑛 {𝑏𝑛+1 − [(𝑏𝑛+2 − 𝑏𝑛+3 ) + (𝑏𝑛+4 − 𝑏𝑛+5 ) + ⋯ ]}
≥ 0 𝑓𝑜𝑟 𝑎𝑙𝑙 𝑛 [{𝑏𝑛 } 𝑖𝑠 𝑚𝑜𝑛𝑜𝑡𝑜𝑛𝑖𝑐𝑎𝑙𝑙𝑦 𝑑𝑒𝑐𝑟𝑒𝑎𝑠𝑖𝑛𝑔]
⇒ |𝑅𝑛 | ≤ 𝑏𝑛+1
Q.E.D.
3
∞
EX: Find the sum of ∑
𝑛=1
0<
(−1)𝑛−1
correct to 3 decimal places.
𝑛!
1
1
<
(𝑛 + 1)! 𝑛!
𝐴𝑙𝑠𝑜,
0<
𝑆𝑖𝑛𝑐𝑒 lim
𝑛→∞
( 𝑏𝑛 =
1
)
𝑛!
∴ 𝑏𝑛+1 ≤ 𝑏𝑛 𝑓𝑜𝑟 𝑎𝑙𝑙 𝑛 ∈ 𝑍 +
1
1
<
𝑛!
𝑛
1
=0 ⇒
𝑛
lim
𝑛→∞
1
=0
𝑛!
𝑏𝑦 𝑡ℎ𝑒 𝑆𝑞𝑢𝑒𝑒𝑧𝑒 𝑇ℎ𝑒𝑜𝑟𝑒𝑚
∴ 𝑡ℎ𝑒 𝑠𝑒𝑟𝑖𝑒𝑠 𝑐𝑜𝑛𝑣𝑒𝑟𝑔𝑒𝑠 𝑏𝑦 𝑡ℎ𝑒 𝐴𝑙𝑡𝑒𝑟𝑛𝑎𝑡𝑖𝑛𝑔 𝑆𝑒𝑟𝑖𝑒𝑠 𝑇𝑒𝑠𝑡
𝑆 =1−
1 1 1
1
1
1
+ −
+
−
+
+⋯
2 6 24 120 720 5040
𝑏7 =
1
1
<
5040
2000
= 0.0005 𝑚𝑎𝑛𝑢𝑎𝑙𝑙𝑦
𝑆6 = 1 −
1 1 1
1
1
+ −
+
−
= 0.631944
2 6 24 120 720
𝐵𝑦 𝐸𝑠𝑡𝑖𝑚𝑎𝑡𝑖𝑜𝑛 𝑇ℎ𝑒𝑜𝑟𝑒𝑚, |𝑆 − 𝑆6 | ≤ 𝑏7
∴ 0.631944 −
1
1
≤ 𝑆 ≤ 0.631944 +
⇒ 0.6317456 ≤ 𝑆 ≤ 0.6321424
5040
5040
𝑆 ≈ 𝑆6 = 0. 632
(3 𝑑𝑒𝑐𝑖𝑚𝑎𝑙 𝑝𝑙𝑎𝑐𝑒𝑠)
∞
EX: Aproximate the sum of ∑
𝑛=1
(−1)𝑛
by its first 5 terms (5th partial sum).
𝑛2
𝐹𝑖𝑛𝑑 𝑡ℎ𝑒 𝑖𝑛𝑡𝑒𝑟𝑣𝑎𝑙 𝑖𝑛 𝑤ℎ𝑖𝑐ℎ 𝑠𝑢𝑚 𝑆 𝑙𝑖𝑒𝑠.
𝑆5 = −1 +
1 1 1
1
− +
−
≈ −0.838611
4 9 16 25
1
1
≤
(𝑛 + 1)2 𝑛2
⇒
lim
𝑛→∞
1
=0
𝑛2
∴ 𝑏𝑛+1 ≤ 𝑏𝑛 𝑓𝑜𝑟 𝑎𝑙𝑙 𝑛 ∈ 𝑍 +
∴ 𝑡ℎ𝑒 𝑠𝑒𝑟𝑖𝑒𝑠 𝑐𝑜𝑛𝑣𝑒𝑟𝑔𝑒𝑠 𝑏𝑦 𝑡ℎ𝑒 𝐴𝑙𝑡𝑒𝑟𝑛𝑎𝑡𝑖𝑛𝑔 𝑆𝑒𝑟𝑖𝑒𝑠 𝑇𝑒𝑠𝑡
𝐸𝑟𝑟𝑜𝑟 𝑏𝑜𝑢𝑛𝑑 𝑖𝑠 𝑔𝑖𝑣𝑒𝑛 𝑏𝑦: |𝑆 − 𝑆5 | = |𝑅5 | ≤
𝑆𝑜 𝑎𝑐𝑡𝑢𝑎𝑙 𝑠𝑢𝑚 𝑙𝑖𝑒𝑠 𝑏𝑒𝑡𝑤𝑒𝑒𝑛
1
≈ 0.027778
36
− 0.838611 − 0.027778 ≤ 𝑆 ≤ −0.838611 + 0.027778
4
ABSOLUTE AND CONDITIONAL CONVERGENCE (for loco series +, - ,+,+,+,-,-,-)
∞
Given 𝐀𝐍𝐘 series
∞
∑ 𝑎𝑛 we can consider the corresponding series ∑|𝑎𝑛 | = |𝑎1 | + |𝑎2 | + ⋯
𝑛=1
𝑛=1
Absolute Convergence
∞
∞
The series ∑ 𝑎𝑛 is 𝐚𝐛𝐬𝐨𝐥𝐮𝐭𝐥𝐲 𝐜𝐨𝐧𝐯𝐞𝐫𝐠𝐞𝐧𝐭 if ∑|𝑎𝑛 | converges.
𝑛=1
𝑛=1
If 𝑎𝑛 ≥ 0 for all n, absolute convergence is the same as convergence.
Conditional Convergence
∞
∞
∞
The series ∑ 𝑎𝑛 is 𝐜𝐨𝐧𝐝𝐢𝐭𝐢𝐨𝐧𝐚𝐥𝐥𝐲 𝐜𝐨𝐧𝐯𝐞𝐫𝐠𝐞𝐧𝐭 if ∑ 𝑎𝑛 converges but ∑|𝑎𝑛 | diverges .
𝑛=1
𝑛=1
𝑛=1
∞
(−1)𝑛−1
𝐸𝑋: ∑
𝑖𝑠 𝑐𝑜𝑛𝑑𝑖𝑡𝑖𝑎𝑙𝑙𝑦 𝑐𝑜𝑛𝑣𝑒𝑟𝑔𝑒𝑛𝑡
𝑛
𝑛=1
Theorem of Absolute Convergence
∞
If ∑ 𝑎𝑛 is absolutely convergent then it is also convergent.
𝑛=1
Proof:
by definition of absolute value: − |𝑎𝑛 | ≤ 𝑎𝑛 ≤ |𝑎𝑛 | ⇒ 0 ≤ 𝑎𝑛 + |𝑎𝑛 | ≤ 2|𝑎𝑛 |
∞
∞
If ∑ 𝑎𝑛 is absolutely convergent then 2 ∑|𝑎𝑛 | is convergent.
𝑛=1
𝑛=1
∴ by the Comparison Test ∑∞
𝑛=1(𝑎𝑛 + |𝑎𝑛 |) is also convergent.
∞
𝑆𝑖𝑛𝑐𝑒 ∑ 𝑎𝑛 𝑖𝑠 𝑡ℎ𝑒 𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑐𝑒 𝑜𝑓 𝑡𝑤𝑜 𝑐𝑜𝑛𝑣𝑒𝑟𝑔𝑒𝑛𝑡 𝑠𝑒𝑟𝑖𝑒𝑠
𝑛=1
∞
∞
∞
∑ 𝑎𝑛 = ∑(𝑎𝑛 + |𝑎𝑛 |) − ∑|𝑎𝑛 |
𝑛=1
𝑛=1
𝑛=1
∞
∴ ∑ 𝑎𝑛 is convergent
𝑛=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 ∑
𝑛=1
∞
∑
𝑛=1
cos 𝑛
is convergent.
𝑛2
cos 𝑛
cos 1 cos 2
=
+ 2 +⋯
2
𝑛
12
2
𝑡𝑒𝑟𝑚𝑠 𝑤𝑖𝑡ℎ 𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑡 𝑠𝑖𝑔𝑛𝑠, ( 𝑛𝑜𝑡 𝑎𝑙𝑡𝑒𝑟𝑛𝑎𝑡𝑖𝑛𝑔 𝑠𝑒𝑟𝑖𝑒𝑠).
∞
cos 𝑛
1
𝐻𝑜𝑤𝑒𝑣𝑒𝑟 | 2 | ≤ 2
𝑛
𝑛
𝑓𝑜𝑟 𝑎𝑙𝑙 𝑛 ∈ 𝑍
+
𝑎𝑛𝑑 ∑
𝑛=1
∞
∴ 𝑏𝑦 𝑡ℎ𝑒 𝐶𝑜𝑚𝑝𝑎𝑟𝑖𝑠𝑜𝑛 𝑇𝑒𝑠𝑡,
∑|
𝑛=1
1
𝑖𝑠 𝑐𝑜𝑛𝑣𝑒𝑟𝑔𝑒𝑛𝑡.
𝑛2
cos 𝑛
| 𝑖𝑠 𝑐𝑜𝑛𝑣𝑒𝑟𝑔𝑒𝑛𝑡,
𝑛2
𝑎𝑛𝑑 𝑏𝑦 𝑡ℎ𝑒 𝑇ℎ𝑒𝑜𝑟𝑒𝑚 𝑜𝑓 𝐴𝑏𝑠𝑜𝑙𝑢𝑡𝑒 𝐶𝑜𝑛𝑣𝑒𝑟𝑔𝑒𝑛𝑐𝑒,
∞
𝑠𝑜 𝑖𝑠
∑
𝑛=1
cos 𝑛
𝑛2
So what is important about absolute and conditional convergence?
Algebra: a + b = b + a. If we have a finite sum ∑ 𝑎𝑛 , 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!
𝐸𝑋:
1 1 1 1 1
𝑆=1− + − + − +⋯
2 3 4 5 6
⇒
1
1 1 1 1
𝑆 = − + − +⋯
2
2 4 6 8
𝑎𝑑𝑑 𝑡ℎ𝑒𝑚 𝑡𝑜𝑔𝑒𝑡ℎ𝑒𝑟
3
1 1 1
2
1 1 1 1 1
𝑆 = 1 + 0 + − + + 0 ⋯ ⇒ 𝑆 = (1 − + − + − + ⋯ )
2
3 2 5
3
2 3 4 5 6
Thus we get a rearrangement of the original series with a different sum!
● If ∑ 𝑎𝑛 is absolutely convergent and its value is s then any rearrangement of ∑ 𝑎𝑛 will also have
a value of s.
● If ∑ 𝑎𝑛 is conditionally convergent and r is any real number then there is a rearrangement of
∑ 𝑎𝑛 whose value will be r. NO KIDDING !!!!!! But the series is convergent !!!!!!!!!! Yet, we don’t
know the sum. (proof shown by Riemann) -
(−1)𝑛−1
▪ We found that alternating harmonic series ∑
is conditionally convergent, so no matter what value
𝑛
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)𝑛
▪ We found that alternating series ∑ 2 is absolutely convergent and so no matter how we rearrange the
𝑛
terms of this series we’ll always get the same value. - . In fact it can be shown that the value of that series
is
∞
∑
𝑛=1
(−1)𝑛
𝜋2
=
−
𝑛2
12
6
ALTERNATING and LOCO SERIES PROBLEMS
1. Determine if each of the following series are absolute convergent, conditionally convergent or divergent.
∞
(−1)𝑛−1
(𝑎) ∑
𝑛
𝑛=1
∞
(−1)𝑛
( 𝑏) ∑
𝑛2
𝑛=1
∞
sin 𝑛
(𝑐) ∑ 3
𝑛
𝑛=1
∞
(−1)𝑛 𝑛2
(𝑑) ∑ 2
𝑛 +5
𝑛=1
∞
(−1)𝑛−3 √𝑛
𝑒) ∑
𝑛+4
𝑛=1
∞
𝑓) ∑
𝑛=2
cos(𝑛𝜋)
√𝑛