Number 8 Is Interesting
Eight is the first (after the trivial 1) cube, 2³ = 8 [Bunch]. Eight is the only cube 1 less than a square, 8 = 3² - 1. [Bunch] Among the Fibonacci numbers there are just so many cubes: u-6 = -8, u-2 = -1, u0 = 0, u-1 = u1 = u2 = 1, u6 = 8 [Bunch, Roberts].
Bryan Bunch's and Joe Roberts' books list many more curiosities with 8 in a prominent role. The former is almost entirely elementary, the latter often refers to more advanced topics from college mathematics. A novel property of 8 has been recently made public by Paul and Vincent Steinfeld of Germany.
Consider the sequence
| (*) |
| |
9876
 1234 |
|
| |
98765
 12345 |
|
| ... |
Steinfelds suggest an interpretation of what the terms that come after one runs out of decimal digits may look like. To boot, define
| | Nn | = ∑nk = 0 (10 - (k + 1)) 10 - k and
|
| | Dn | = ∑nk = 0 (k + 1) 10 - k. |
Then, with a suitable placement of the decimal point, the ratio Rn = Nn / Dn (for small values of n) is exactly the generic term of the above sequence.
The limits limn→∞Nn and limn→∞Dn are easily computed to be
| | limn→∞Nn | = 800/81 and
|
| | limn→∞Dn | = 100/81. |
So that limn→∞Rn = 8. Unexpectedly, the sequence (*) converges to 8!
References
- B. Bunch, The Kingdom of Infinite Number: A Field Guide, W. H. Freeman & Co., 2000
- J. Roberts, Lure of the Integers, MAA, 1992
- P. and V. Steinfeld, Math Bite: A Magic Eight, Mathematics Magazine, Vol. 82, No. 1, Feb. 2009, p. 25
Copyright © 1996-2009 Alexander Bogomolny
First we are going to show that limn→∞Dn = 100/81. Note that the series at hand is absolutely convergent (it's convergent and consists of positive terms), meaning that it is possible to reshuffle the terms without affecting the limit. With the formula for the sum of the geometric series in mind, we have
| | ∑nk = 0 (k + 1) 10 - k | = ∑nk = 0 10 - k | + ∑nk = 1 k 10 - k | | |
| | | = ∑nk = 0 10 - k | + ∑nk = 1 10 - k | + ∑nk = 2 10 - k | + ∑nk = 3 10 - k ... |
| | | = 10/9 | + 1/9 | + 10-1 /9 | + 10-2 /9 ... |
| | | = 10/9 (1 | + 10-1 | + 10-2 | ...) |
| | | = (10/9)² | = 100/81, | |
as claimed.
For the limit limn→∞Nn, observe that the series is again absolutely convergent (the series of absolute values is convergent), and
| | Nn = 10 ∑nk = 0 10 - k - Dn, |
such that in the limit we get
| | limn→∞Nn | = 10×10/9 - 100/81 |
| | | = 900/81 - 100/81 |
| | | = 800/81. |
Copyright © 1996-2009 Alexander Bogomolny
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