[jbjb]

Originally Posted by pdurrant

Here's the answer:

Spoiler:

It's sometimes easier to look at fractions rather than percentages. With the conditions in this problem, every second from the start of the race, Achilles covers

1/10, 1/20, 1/30, 1/40, 1/50, 1/60, 1/70, 1/80, etc.

So to find out if Achilles finishes the race, we need to know if the sum of all those fractions ever reaches 1.

Consider the following sequence of fractions:

1/10, 1/20, 1/20, 1/40, 1/40, 1/40, 1/40, 1/80, 1/80, 1/80, ....

Each of these fractions is less than or equal to the corresponding fraction in the sequence we're interested in.

Spoiler:

I don't think that explanation of the diverging sum is quite right. For example, if you take the third item in each sequence, the entry from the second sequence is not actually less than or equal to that from the first.

Perhaps a better way would be to group them as:
1/10, 1/20, (1/30 + 1/40), (1/50 + 1/60 + 1/70 + 1/80) ... etc.

In general, the nth group of terms (where n starts at 1 for (1/30+1/40)) will (for all n > 0) be a sum of 2^n fractions, the least of which will be 1/(20*(2^n)). Thus the sum of each group must be at least (2^n)/(20*(2^n)), or 1/20. A sum of an infinite sequence of groups, each of which has a sum of at least 1/20 will itself be infinite.

Cheers,
John