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Let $S\subseteq \mathbb Z^+$ be a set of positive odd numbers. I am asked to prove that there exists a sequence $(x_n)$ such that for any positive integer $k$, $$ \sum_{n=1}^\infty x_n^k $$ converges iff $k\in S$. I have no idea where to start. Even in the special case $S=\{1\}$ I don't know if any sequence would work.

Any hints?

If $S=\{1\}$, then we have to find a sequence such that $\sum x_n$ converges but $\sum x_n^k$ doesn't for $k\geq 2$. However, for a positive sequence, if $\sum x_n^k$ converges then $\sum x_n^{k+1}$ converges, so we must not choose $(x_n)$ to be a sequence of positive terms. But what can I do next?

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For $S=\{1\}$, one can use a sequence of the form $$ 2c_1,-c_1,-c_1;2c_2,-c_2,-c_2;2c_3,-c_3,-c_3,\dots $$ where $\{c_1,c_2,\dots\}$ is, say, a sequence decreasing to $0$ not too fast, such as $c_j = 1/\log(j+1)$.

For more complicated $S$, I suspect a similar approach will work, using this sort of "hybridization" of a periodic sequence whose appropriate powers sum to $0$ over every period with a slowly decreasing sequence.

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