A rubber band is 1 meter long. An ant starts at one end, crawling at 1 millimeter per second. At the end of each second, the rubber band is instantaneously stretched by an additional meter. (So, at the end of the nth second, the rubber band becomes n+1 meters long.)

Does the ant ever reach the far end of the band? If so, when?

(In reply to re(2): The backward ant by Aeternus)

Thanks, Aeturnus. That helped a lot!

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Call D(t) the length of the rubber band after t seconds.
D(0) = 1m = 1000mm
D(1) = 2m = 2000mm
D(2) = 3m = 3000mm
In general:
D(t) = (t + 1)m = 1000(t + 1)mm

Call d(t) the distance between the ant and the beginning after t seconds.
d(0) = 0mm
d(1) = 1(2/1)mm
d(2) = [1(2/1) + 1](3/2) = 1(2/1)(3/2) +1(3/2) = 1(3/1) + 1(3/2)
d(3) = [1(3/1) + 1(3/2) +1](4/3) = 1(3/1)(4/3) + 1(3/2)(4/3) + 1(4/3) = 1(4/1) + 1(4/2) + 1(4/3)
In general:
d(t) = 1[(t +1)/1] + 1[(t + 1)/2] + 1[(t + 1)/3] + ... + 1[(t + 1)/t]

   t
=  ∑ (t = 1)/i
  i=1

The ant reaches the end at the earliest time T where

d(T) ≥ D(T)

  T
  ∑(T +1)/i ≥ 1000(T +1)
i=1

(T + 1)∑ (T +1)/i ≥ (T +1)1000

1000 ≤ ∑1/i = H(T), where we recognize H(T) as the first T terms in the harmonic series.

Based on Oresme's proof of the divergence, we know that H(2^n) > (n + 2)/2 for n>2.

If we let (n + 2)/2 = 1000, then n= 1998 and T ≤ 2^1998.

We can narrow the range further using the Euler-Maclaurin summation formula(see Aeternus' last post), but that is a little too much at this time.

Posted by TomM on 2002-10-05 11:37:01