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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.

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  • $\begingroup$ Can you provide a source for this question? $\endgroup$ – Tad May 9 '16 at 14:33
  • $\begingroup$ @Tad Maths Tripos Analysis I example sheet 1, question 16. $\endgroup$ – Columbo May 9 '16 at 14:48
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Lemma 1 For any finite set $S$ of odd positive integers there is a finite multiset $A$ or real numbers such that $\sum_{\alpha \in A} \alpha^k = 0$ iff $k\in S$.

Proof For $S = \{k\}$ take $A_k = [1,1,(-2)^{1/k}]$.

Let $S = \{k_1,\dots, k_m\}$. Put $A = \big[\prod_{j=1}^m \alpha_j\mid \alpha_j\in A_{k_j} \text{ for } j=1,\dots,m\big]$. Then $$ \sum_{\alpha \in A} \alpha^k = \sum_{\alpha_1\in A_{k_1}}\dots \sum_{\alpha_m\in A_{k_m}}\prod_{j=1}^m \alpha_j^k = \prod_{j=1}^m \sum_{\alpha_j\in A_{k_j}}\alpha_j^k, $$ which is equal to zero iff one of multiplicands $\sum_{\alpha_j\in A_{k_j}}\alpha_j^k$ is zero, i.e. iff $k=k_j$ for some $j$. $\Box$

Let first $S$ be finite, and $A$ be the multiset corresponding to $S$; $|A|= m$.

Take a positive sequence $a_n$ such that $\sum_{n=1}^\infty a^k_n = \infty$ for each $k\ge 1$ (e.g. $a_n = (\log n+1)^{-1}$) and define $x_{(n-1)m+1},\dots,x_{nm}$ to be equal to $\alpha a_n$, $\alpha\in A$, in some order. Obviously, this satisfies the requirement.

Lemma 2 For any odd positive integer $m$ there is a sequence $\{x_n,n\ge 1\}$ or real numbers such that the sequence $\{z_1^m+\dots+z_n^m,n\ge 1\}$ is unbounded, while $\sup_{k\neq m,n\ge 1}|z_1^k+\dots+z_n^k|<\infty$.

Proof Use the above construction with $S=\{1,2,\dots,m-1\}$, $a_n = n^{-1/m}$, to get $\{z_n,n\ge 1\}$. Then $\{z_1^k+\dots+z_n^k,n\ge 1\}$ is obviously bounded for $k<m$, unbounded for $k=m$, and $$\sup_{k> m,n\ge 1}|z_1^k+\dots+z_n^k| \le C \sum_{n=1}^{\infty} n^{-(m+1)/m}<\infty, $$ as required. $\Box$

Now let $S$ be arbitrary set of odd positive integers, $T$ be its complement and $\{k_n,n\ge 1\}$ be a sequence of integers from $T$ such that each integer from $T$ appears in it infinitely often.

Using Lemma 2, for each $n\ge 1$, we can construct a sequence $\{y_j(n),j=1,\dots,m_n\}$ such that $\sum_{j=1}^{m_n} y_j(n)^{k_n}\ge 1$ and $\sup_{k\neq k_n,j=1,\dots,m_n}|y_1(n)^k+\dots+y_j(n)^k|<2^{-n}$. Setting \begin{align} & x_i = y_i(1), &1\le i\le m_1,\\ & x_i = y_{i-m_1}(2), &m_1<i\le m_1+m_2,\\ & x_i = y_{i-m_1-m_2}(3), &m_1+m_2<i\le m_1+m_2+m_3 \end{align} etc, we get the required series. Moreover, $|\sum_{n=1}^\infty x_n^k|\le 1$ for $k\in S$, and the series converges uniformly in $k\in S$.

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  • $\begingroup$ Yep, that looks right. My supervisor pointed me towards a magazine article that covered it in a less concise fashion, but using a product and exploiting zero factors is very clever indeed! $\endgroup$ – Columbo May 11 '16 at 22:10
  • $\begingroup$ The idea of zeros I've borrowed from @zhw.'s answer; it is very simple and clever indeed. I doubt that I would be able to solve the problem without it. $\endgroup$ – zhoraster May 12 '16 at 4:31
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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.

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