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

(*)

987

123

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

	N_n	= ∑ⁿ_{k = 0} (10 - (k + 1)) 10^{- k} and
	D_n	= ∑ⁿ_{k = 0} (k + 1) 10^{- k}.

Then, with a suitable placement of the decimal point, the ratio R_n = N_n / D_n (for small values of n) is exactly the generic term of the above sequence.

The limits lim_n→∞N_n and lim_n→∞D_n are easily computed to be

	lim_n→∞N_n	= 800/81 and
	lim_n→∞D_n	= 100/81.

So that lim_n→∞R_n = 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

First we are going to show that lim_n→∞D_n = 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

∑^∞_{k = 0} (k + 1) 10^{- k}	= ∑^∞_{k = 0} 10^{- k}	+ ∑^∞_{k = 1} k 10^{- k}
	= ∑^∞_{k = 0} 10^{- k}	+ ∑^∞_{k = 1} 10^{- k}	+ ∑^∞_{k = 2} 10^{- k}	+ ∑^∞_{k = 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 lim_n→∞N_n, observe that the series is again absolutely convergent (the series of absolute values is convergent), and

N_n = 10 ∑^∞_{k = 0} 10^{- k} - D_n,

such that in the limit we get

	lim_n→∞N_n	= 10×10/9 - 100/81
		= 900/81 - 100/81
		= 800/81.

Note: You may want to check that, in base b > 2, the role of 8 is taken by b - 2.

Number Curiosities

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