IMC 2013 · Day 1 · P5

The answer is YES. We prove a more general statement; suppose that $\mathbb{N} = C \cup D$ is an arbitrary decomposition of $\mathbb{N}$ 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 $D_k = D \cap [1, k]$.

Lemma. For every positive integer $k$ there exists a positive integer $N_k$ and a sequence $X_k = (x_{k,1}, \dots, x_{k,N_k})$ of complex numbers with the following properties:

(a) For $p \in D_k$, we have $\left| \sum\limits_{j=1}^{N_k} x_{k,j}^p \right| \ge 1$.

(b) For $p \in C_k$, we have $\sum\limits_{j=1}^{N_k} x_{k,j}^p = 0$; moreover, $\left| \sum\limits_{j=1}^{m} x_{k,j}^p \right| \le \frac{1}{k}$ holds for $1 \le m \le N_k$.

Proof. First we find some complex numbers $z_1, \dots, z_k$ with $$\tag{1} \sum_{j=1}^{k} z_j^p = \begin{cases} 0 & p \in C_k \\ 1 & p \in D_k \end{cases}$$ As is well-known, this system of equations is equivalent to another system $\sigma_\nu(z_1, \dots, z_k) = w_\nu$ ($\nu = 1, 2, \dots, k$) 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 $z_1, \dots, z_k$ are the roots of the polynomial $z^k - w_1 z^{k-1} + - \dots + (-1)^k w_k$ in some order.

Now let $$M = \left\lceil \max_{1 \le m \le k,\ p \in C_k} \left| \sum_{j=1}^{m} z_j^p \right| \right\rceil$$ and let $N_k = k \cdot (kM)^k$. We define the numbers $x_{k,1}, \dots, x_{k,N_k}$ by repeating the sequence $\left( \frac{z_1}{kM}, \frac{z_2}{kM}, \dots, \frac{z_k}{kM} \right)$ $(kM)^k$ times, i.e.\ $x_{k,\ell} = \frac{z_j}{kM}$ if $\ell \equiv j \pmod{k}$. Then we have $$\sum_{j=1}^{N_k} x_{k,j}^p = (kM)^k \sum_{j=1}^{k} \left( \frac{z_j}{kM} \right)^p = (kM)^{k-p} \sum_{j=1}^{k} z_j^p;$$ 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 $1 \le m \le N_k$; then $m = kr + s$ with some integers $r$ and $1 \le s \le k$ and hence $$\left| \sum_{j=1}^{m} x_{k,j}^p \right| = \left| \sum_{j=1}^{kr} + \sum_{j=kr+1}^{kr+s} \right| = \left| \sum_{j=1}^{s} \left( \frac{z_j}{kM} \right)^p \right| \le \frac{M}{(kM)^p} \le \frac{1}{k}.$$ The lemma is proved.

Now let $S_k = N_1 + \dots + N_k$ (we also define $S_0 = 0$). Define the sequence $(a_n)$ by simply concatenating the sequences $X_1, X_2, \dots$: $$(a_1, a_2, \dots) = (x_{1,1}, \dots, x_{1,N_1}, x_{2,1}, \dots, x_{2,N_2}, \dots, x_{k,1}, \dots, x_{k,N_k}, \dots);$$ $$a_{S_k + j} = x_{k+1, j} \quad (1 \le j \le N_{k+1}).$$ If $p \in D$ and $k \ge p$ then $$\left| \sum_{j=S_k+1}^{S_{k+1}} a_j^p \right| = \left| \sum_{j=1}^{N_{k+1}} x_{k+1,j}^p \right| \ge 1;$$ By Cauchy's convergence criterion it follows that $\sum a_n^p$ is divergent.

If $p \in C$ and $S_{u-1} < n \le S_u$ with some $u \ge p$ then $$\left| \sum_{n=S_p+1}^{n} a_n^p \right| = \left| \sum_{k=p+1}^{u-1} \sum_{j=1}^{N_k} x_{k,j}^p + \sum_{j=1}^{n-S_{u-1}} x_{u,j}^p \right| = \left| \sum_{j=1}^{n-S_{u-1}} x_{u,j}^p \right| \le \frac{1}{u}.$$ Then it follows that $\sum\limits_{n=S_p+1}^{\infty} a_n^p = 0$, and thus $\sum\limits_{n=1}^{\infty} a_n^p = 0$ is convergent.