I thought that this question is more suitable for MSE, and asked it there. (Link to the MSE question) However, it does not get any answer despite the upvotes. It appears that I might have underestimated its difficulty. Please tell me if it is suitable for MO; ask me to delete it if it is inappropriate.

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 if $k\in S$ and diverges if $k\notin S$.

I have no idea where to start. In the special case $S=\{1\}$, we can let the sequence $(x_n)$ be $$ y_1,y_1,-2y^1,y_2,y_2,-2y_2,\ldots, $$ where $y_n\to 0$ but $y_n$ decreases very slowly. However I doubt if this would work for more complicated $S$. Note that $S$ can be infinite.

Any hints?

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    $\begingroup$ From the way you ask, this sounds like a homework question (which as you guessed is discouraged here). I will suggest a hint (which I haven't completely worked out, but I think will work). For each odd number, $2k+1$, here's a suggested way to build a sequence which diverges for that power alone. Find a finite collection of numbers $T_k=\{a_1,\ldots,a_l\}$ such that $\sum_{a\in T_k}a^{2j+1}=0$ for $j=1,\ldots,k-1$ and $\sum_{a\in T_k}a^{2j+1}\ne 0$ for $j\ge k$ (e.g. Prouhet). Then build a sequence by concatenating the block $a_12^{-j},\ldots,a_l2^{-j}$ repeated $2^{(2k+1)j}$ times over. $\endgroup$ – Anthony Quas Oct 2 '19 at 10:41
  • $\begingroup$ This guarantees convergence (to 0) for every odd power below $2k+1$; divergence for $2k+1$ and convergence for everything above. Now find a way to mix these sequences. $\endgroup$ – Anthony Quas Oct 2 '19 at 10:42
  • $\begingroup$ "I am asked to prove..." Who asked you to prove this, and under what circumstances, please? $\endgroup$ – Gerry Myerson Oct 2 '19 at 13:21
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    $\begingroup$ @GerryMyerson I cannot remember exactly. I saw it about a year ago on some website, which I cannot remember. It is NOT my homework. $\endgroup$ – Ma Joad Oct 2 '19 at 13:34
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    $\begingroup$ So, on some website, you were personally asked to prove this? $\endgroup$ – Gerry Myerson Oct 10 '19 at 9:57

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