International Mathematics Competition

The answer is YES. We prove a more general statement; suppose that $N=C \cup D$ is an arbitrary decomposition of $N$ into two disjoint sets. Then there exists a sequence $(a_{n})_{n=1}^{\infty}$ such that $\sum_{n=1}^{\infty} a_{n}^{p}$ is convergent for $p \in C$ and divergent for $p \in D$. Define $C_{k}=C \cap[1, k]$ and $D_{k} \cap[1, k]$. Lemma. For every positive integer $k$ there exists a positive integer $N_{k}$ and a sequence $X_{k}=(x_{k, 1}, \ldots, x_{k, N_{k}})$ of complex numbers with the following properties: (a) For $p \in D_{k}$, we have $|\sum_{j=1}^{N_{k}} x_{k, j}^{p}| \geq 1$. (b) For $p \in C_{k}$, we have $\sum_{j=1}^{N_{k}} x_{k, j}^{p}=0$; moreover, $|\sum_{j=1}^{m} x_{k, j}^{p}| \leq \frac{1}{k}$ holds for $1 \leq m \leq N_{k}$. Proof. First we find some complex numbers $z_{1} \ldots, z_{k}$ with $\sum_{j=1}^{k} z_{j}^{p}= \begin{cases}0 & p \in C_{k} \\ 1 & p \in D_{k}\end{cases}$. As is well-known, this system of equations is equivalent to another system $\sigma_{\nu}(z_{1}, \ldots, z_{k})=w_{\nu}(\nu= 1,2, \ldots, k)$ where $\sigma_{\nu}$ is the $\nu$ th elementary symmetric polynomial, and the constants $w_{\nu}$ are uniquely determined by the Newton-Waring-Girard formulas. Then the numbers $z_{1}, \ldots, z_{k}$ are the roots of the polynomial $z^{k}-w_{1} z^{k-1}+-\ldots+(-1)^{k} w_{k}$ in some order. Now let $M=\lceil\max_{1 \leq m \leq k, p \in C_{k}}|\sum_{j=1}^{m} z_{j}^{p}|\rceil$ and let $N_{k}=k \cdot(k M)^{k}$. We define the numbers $x_{k, 1} \ldots, x_{k, N_{k}}$ by repeating the sequence $(\frac{z_{1}}{k M}, \frac{z_{2}}{k M}, \ldots, \frac{z_{k}}{k M})$ $(k M)^{k}$ times, i.e. $x_{k, \ell}=\frac{z_{j}}{k M}$ if $\ell \equiv j(\bmod k)$. Then we have $\sum_{j=1}^{N_{k}} x_{k, j}^{p}=(k M)^{k} \sum_{j=1}^{k}(\frac{z_{j}}{k M})^{p}=(k M)^{k-p} \sum_{j=1}^{k} z_{j}^{p}$ then from (1) the properties (a) and the first part of (b) follows immediately. For the second part of (b), suppose that $p \in C_{k}$ and $1 \leq m \leq N_{k}$; then $m=k r+s$ with some integers $r$ and $1 \leq s \leq k$ and hence $|\sum_{j=1}^{m} x_{k, j}^{p}|=|\sum_{j=1}^{k r}+\sum_{j=k r+1}^{k r+s}|=|\sum_{j=1}^{s}(\frac{z_{j}}{k M})^{p}| \leq \frac{M}{(k M)^{p}} \leq \frac{1}{k}$. The lemma is proved. Now let $S_{k}=N_{1} \ldots, N_{k}$ (we also define $S_{0}=0$ ). Define the sequence (a) by simply concatenating the sequences $X_{1}, X_{2}, \ldots$: $(a_{1}, a_{2}, \ldots)=(x_{1,1}, \ldots, x_{1, N_{1}}, x_{2,1}, \ldots, x_{2, N_{2}}, \ldots, x_{k, 1}, \ldots, x_{k, N_{k}}, \ldots)$ $a_{S_{k}+j}=x_{k+1, j} \quad(1 \leq j \leq N_{k+1})$. If $p \in D$ and $k \geq p$ then $|\sum_{j=S_{k}+1}^{S_{k+1}} a_{j}^{p}|=|\sum_{j=1}^{N_{k+1}} x_{k+1, j}^{p}| \geq 1$. By Cauchy's convergence criterion it follows that $\sum a_{n}^{p}$ is divergent. If $p \in C$ and $S_{u}<n \leq S_{u+1}$ with some $u \geq p$ then $|\sum_{j=S_{p}+1}^{n} a_{n}^{p}|=|\sum_{k=p+1}^{u-1} \sum_{j=1}^{N_{k}} x_{k, j}^{p}+\sum_{j=1}^{n-S_{u-1}} x_{u, j}^{p}|=|\sum_{j=1}^{n-S_{u-1}} x_{u, j}^{p}| \leq \frac{1}{u}$. Then it follows that $\sum_{n=S_{p}+1}^{\infty} a_{n}^{p}=0$, and thus $\sum_{n=1}^{\infty} a_{n}^{p}=0$ is convergent.