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# Sequence such that series of kth powers converges for specific k watch

1. Question:

Given a set S of odd positive integers, prove there is a sequence such that converges if and only if k is in S.

Solution (so far):

I think I have proved this for any finite S.
Notation; let . Then but diverges for any power k. So I use this sequence to give the asymptotic size of terms , and arrange the coefficients to have cancellation if and only if k is in S.

E.g. when (only one element), I take , and .
Then , and then the partial sum for all M, so we have convergence when we need it.

And for we have , where K is non-zero and independent of n, and thus so we have the divergence when we need it.

Similar construction for any finite S, can arrange cancellation for only powers k in S.

However I'm stuck on the case of infinite S.

Apart from obviously if S = {all odd numbers} then an alternating sequence like will work. But for a general infinite set S, I can't think of the construction. Can the above ideas be adapted or is a new approach needed?
2. I'm not sure about an explicit construction, but you can derive the answer for infinite S from finite S using compactness: consider each x_n to lie in the compact interval [-1,1]. By Tychonoff's theorem, the set of sequences (x_n) is also compact.

Now take a sequence of nested finite sets S_i which converge to your (infinite) S - these have the finite intersection property, and hence the intersection of all of them is non-empty. Clearly, any sequence in their intersection will do.
3. But any sequence x_n can satisfy the required criteria for at most one set S_i, since the series must converge if and only if the power k is in S_i.
4. the question doesn't ask that the set s be infinite.
5. But it doesn't say finite either, so we can't assume either, and must be able to do both to have properly answered it.
6. Does anyone have any other ideas on this?

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Updated: April 1, 2011
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