Conditionally convergent power sums I'm struggling on the following question:

Let $S$ be a (possibly infinite) set of odd positive integers. Prove that
  there exists a real sequence $(x_n)$ such that, for each positive integer $k$, the
  series $\sum x_n^k$ converges iff $k \in S$.

I'm completely lost on this one. How can we even form a sequence such that the series converges for $k = 3, 7$ but not $5$? The series are all conditionally convergent, perhaps some clever rearrangement of the alternating harmonic series could do it.

On some questions you raised: No careful rearrangement $(x_n)$ of the harmonic alternating series can work, because in that case $\sum x_n^k$ converges absolutely for all $k\ge 2.$ As for the weaker problem of finding $(x_n)$ such that $\sum x_n^k$ converges for $k=3,7,$ but not for $k=5,$ we can do this: Define $x_n$ three terms at a time so that the $m$th triple is
$$\frac{1}{m^{1/5}},\frac{1}{m^{1/5}} , \frac{-2^{1/3}}{m^{1/5}}.$$
Summing $x_n^3$ over the $m$th triple gives $0.$ Hence $\sum x_n^3$ converges. But summing $x_n^5$ over the $m$th triple gives $(2-2^{5/3})/m.$ Thus $\sum x_n^5 = -\infty.$ Because $7/5>1,$ we see $\sum x_n^7$ converges absolutely. 
