1/1000 + 1/2000 + ... + 1/1000n >= 1.
It is a theorem that for all n >= 1,
1 + 1/2 + 1/3 + ... + 1/n = log n + gamma + q(n),
where 0 <= q(n) <= 1/2n and gamma (`Euler's constant') is 0.5772156649...
We can conclude that our ant will take around
e^(1000-gamma) ~= 1.1 * 10^434 seconds
to reach the end of the string.
By the way, 1 m/s is a pretty fast ant!
David Moews ( email@example.com )
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