John Malinckroft's explanation (see below) was clear to me. He is not
only a good mathematician, he is a very good teacher. A teacher should
be able to guess the background of his typical students, not to teach
over their heads and blame them for not being at the expected level.
I am not criticizing those who overestimated my level of mathematical
sophistication. I thank Uri and Jack for contributing to my apprciation
of nuances.
Please respond to my "head counting". What can be gained from it? I do
not know. Just another set of statistical data to think about.
What follows is John's text. It shows how the wording of the problem
naturally leads to a harmonic series.
Ludwik Kowalski
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I think the simplest way to solve the discrete problem is to see that in
the first second the worm covers a fraction equal to 1/10^5 of the rope's
length. In the second second it covers 1/(2*10^5) of the rope's length
(since the rope is now twice as long.) In the third second it covers
1/(3*10^5) of the rope's length, and so on. We need all these fractions
to add up to 1 (the *whole* rope) so we must find n such that
(1/10^5)*(1/1 + 1/2 + 1/3 + ... + 1/n) = 1
or
sum from k = 1 to n of 1/k = 10^5
To solve the problem it is enough to understand the definition of Euler's
constant (gamma = .577..) which is
gamma = limit as n -> infinity of (sum from k = 1 to n of 1/k - ln(n))
which means that, for large enough n,
sum from k = 1 to n of 1/k ~= ln(n) + gamma
Using this result we have
ln(n) + gamma ~= 10^5
so n ~= exp(10^5 - gamma)
Since n is indeed *very* large, our assumption is likely justified. (In
fact I can show that it is *extremely* well justified.)
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