Conditionally Convergent:
What’s the big deal?
• Riemann proved that:
X
If
an is a conditionally convergent series and r is any real number whatsoever,
X
then there is a rearrangement of
an that has a sum equal to r.
• That’s VERY WEIRD! Let’s collect some data using the alternating harmonic.
1.
∞
X
(−1)n−1
n=1
1
1 1 1 1 1 1 1 1
1
1
1
1
1
1
=1− + − + − + − + −
+
−
+
−
+
···
n
2 3 4 5 6 7 8 9 10 11 12 13 14 15
Overestimates:
s1 = 1
s3 = 1 − 12 + 13 ≈ .833333
s5 = 1 − 21 + 13 − 41 + 15 ≈ .783333
s7 = 1 − 21 + 13 − 41 + 15 − 16 + 71
≈ .759524
s9 ≈ .745635
s19 ≈ .718771
s29 ≈ .710091
s49 ≈ .703247
s99 ≈ .698172
s199 ≈ .695653
s499 ≈ .694148
s999 ≈ .693647
Underestimates:
s2 = 1 − 21 = .5
s4 = 1 − 12 + 13 − 14 ≈ .583333
s6 = 1 − 12 + 13 − 14 + 51 − 61 ≈ .616667
s8 = 1 − 12 + 13 − 14 + 51 − 61 + 17 − 18
≈ .634524
s10 ≈ .645635
s20 ≈ .668771
s30 ≈ .676758
s50 ≈ .683247
s100 ≈ .688172
s200 ≈ .690653
s500 ≈ .692148
s1000 ≈ .692647
Note: The sum appears to be s ≈ 0.693147. It is actually s = ln 2.
2. Change order: = 1 −
1 1 1 1 1 1 1
1
1
1
1
1
1
1
+ + − + − + +
− +
−
+
+
−
···
2 3 5 4 7 6 9 11 8 13 10 15 17 12
s1 = 1
s2 = 1 − 21 = .5
s5 = 1 − 21 + 13 + 51 − 41 ≈ .783333
s10 = 1 − 21 + 31 + 51 − 14 + 17 − 16 +
s20 ≈ .865424
s30 ≈ .875400
s50 ≈ .883507
s100 ≈ .889662
1
9
+
1
11
−
1
8
≈ .836544
s250 ≈ .893385
s500 ≈ .894631
s1000 ≈ .895255
s2500 ≈ .895630
s5000 ≈ .895755
s7500 ≈ .895796
s10,000 ≈ .895817
s25,000 ≈ .895855
Note: This is definitely a different sum. This series was created to have a sum of 0.9.
•
WOW!!
Absolutely Convergent:
Is this better?
• Let’s collect some data using an absolutely convergent, alternating p-series with p = 2.
1.
∞
X
n=1
(−1)n−1
1
1
1
1
1
1
1
1
1
1
1
1
1
= 1− 2 + 2 − 2 + 2 − 2 + 2 − 2 + 2 − 2 + 2 − 2 + 2 ···
2
n
2
3
4
5
6
7
8
9
10
11
12
13
Overestimates:
s1 = 1
s3 = 1 − 212 + 312 ≈ .861111
s5 = 1 − 212 + 312 − 412 + 512 ≈ .838611
s7 = 1 − 212 + 312 − 412 + 512 − 612 + 712
≈ .831241
s9 ≈ .827962
s19 ≈ .823779
s29 ≈ .823041
s49 ≈ .822671
s99 ≈ .822518
s499 ≈ .822469
s999 ≈ .822468
Underestimates:
s2 = 1 − 212 = .75
s4 = 1 − 212 + 312 − 412 ≈ .798611
s6 = 1 − 212 + 312 − 412 + 512 − 612 ≈ .810833
s8 = 1 − 212 + 312 − 412 + 512 − 612 + 712 − 812
≈ .815616
s10 ≈ .817962
s20 ≈ .821279
s30 ≈ .821930
s50 ≈ .822271
s100 ≈ .822418
s500 ≈ .822465
s1000 ≈ .822467
Note: This sum appears to be s ≈ 0.8224675. It is actually s =
2. Change order: = 1 −
π2
.
12
1
1
1
1
1
1
1
1
1
1
1
+
+
−
+
−
+
+
−
+
−
···
22 32 52 42 72 62 92 112 82 132 102
s1 = 1
s2 = 1 − 212 = .75
s5 = 1 − 212 + 312 + 512 − 412 ≈ .838611
s10 = 1 − 212 + 312 + 512 − 412 + 712 − 612 +
s20 ≈ .831024
s30 ≈ .828571
s50 ≈ .826327
s100 ≈ .824473
1
92
+
1
112
−
1
82
≈ .836227
s250 ≈ .823288
s500 ≈ .822881
s1,000 ≈ .822674
s2,500 ≈ .822550
s5,000 ≈ .822509
s7,500 ≈ .822495
s10,000 ≈ .822488
s25,000 ≈ .822475
Note: Though it takes a long time, it appears this sum is the same.
•
For finite sums, addition is commutative! But,
only absolutely convergent series (infinite sums)
obey that rule!!