Problem

You have 1/2 < a < 1 and the infinite sequence a_0, a_1, a_2 ...
with a_i = a^i.

Can you construct sets S_n, T_n such that for all i < n either a_{2i} \elem S and a_{2i+1} \elem T or a_{2i} \elem T and a_{2i+1} holds, with lim n -> \infty S_n - T_n = 0?
That is to say, the sum of the elements in S_{\infty} - the sum of the elements in T_{\infty} = 0.

Probably.

I'm pretty sure.  To start, put a_1 in S and a_2 in T.  Then step by step assign the larger of each subsequent pair to whichever set has the smaller cumulative sum.

cba to prove it right now, but I think that'll work.

>>3 No guarantee that the difference converges to 0.

>>4
Yes there is.  Once one set "catches up" with the other one, the difference is always smaller than the last term you put in.  The only thing you really need to prove is that it always does catch up.

So say you start off with

S = {1}
T = {a}

so sum(S) > sum(T).  Now add elements in pairs, putting the bigger in T, until sum(T) is larger:

S = {1,a^3,a^5,...,a^(2k+1)}
T = {a,a^2,a^4,...,a^2k}

You'll need to keep going until k is large enough that

a+a^2 (1-a^2k)/(1-a^2) > 1 + a^3 (1-a^2k)/(1-a^2)

(a^2-a^3) (1-a^2k)/(1-a^2) > (1-a)

a^2 (1-a^2k)/(1-a^2) > 1

Since (1-a^2k) tends to 1 for large k, as long as a is big enough that

a^2/(1-a^2) > 1

we can always find such a k.  If a > 1/sqrt(2), you can, but if .5 < a < 1/sqrt(2), T will never catch up and there are no such sequences (Are you sure you copied the problem right? Maybe I made a mistake somewhere.)

Furthermore, since S was ahead of T until the last step, sum(T)-sum(S) at this stage is smaller than a^2k-a^(2k+1).

Now start adding the bigger numbers into S instead of T, and keep going until S catches up again, and all you need to prove now is that it actually does (Hint: Use that estimate of sum(T)-sum(S))