Putnam 2010, Problem B-1

Is there an infinite sequence of real numbers \( a_1, a_2, a_3, \ldots \) such that
\[
a_1^m + a_2^m + a_3^m + \cdots = m
\]
for every positive integer \( m \)?

Solution:
**First solution.** No such sequence exists. If it did, then the Cauchy-Schwarz inequality would imply
$$8 = (a_1^2 + a_2^2 + \cdots)(a_1^4 + a_2^4 + \cdots)$$
$$\geq (a_1^3 + a_2^3 + \cdots)^2 = 9,$$
contradiction.

**Second solution.** (Communicated by Catalin Zara.) Suppose that such a sequence exists. If $a_k^2 \in [0,1]$ for all $k$, then $a_k^4 \leq a_k^2$ for all $k$, and so
$$4 = a_1^4 + a_2^4 + \cdots \leq a_1^2 + a_2^2 + \cdots = 2,$$
contradiction. There thus exists a positive integer $k$ for which $a_k^2 \geq 1$. However, in this case, for $m$ large, $a_k^{2m} > 2m$ and so $a_1^{2m} + a_2^{2m} + \cdots \neq 2m$.

**Third solution.** We generalize the second solution to show that for any positive integer $k$, it is impossible for a sequence $a_1, a_2, \ldots$ of complex numbers to satisfy the given conditions in case the series $a_1^k + a_2^k + \cdots$ converges absolutely. This includes the original problem by taking $k = 2$, in which case the series $a_1^2 + a_2^2 + \cdots$ consists of nonnegative real numbers and so converges absolutely if it converges at all.

Since the sum $\sum_{i=1}^\infty |a_i|^k$ converges by hypothesis, we can find a positive integer $n$ such that $\sum_{i=n+1}^\infty |a_i|^k < 1$. For each positive integer $d$, we then have
$$\left|kd - \sum_{i=1}^n a_i^{kd}\right| \leq \sum_{i=n+1}^\infty |a_i|^{kd} < 1.$$

We thus cannot have $|a_1|, \ldots, |a_n| \leq 1$, or else the sum $\sum_{i=1}^n a_i^{kd}$ would be bounded in absolute value by $n$ independently of $d$. But if we put $r = \max\{|a_1|, \ldots, |a_n|\} > 1$, we obtain another contradiction because for any $\varepsilon > 0$,
$$\limsup_{d \to \infty} (r - \varepsilon)^{-kd} \left|\sum_{i=1}^n a_i^{kd}\right| > 0.$$

For instance, this follows from applying the root test to the rational function
$$\sum_{i=1}^n \frac{1}{1 - a_i^k z} = \sum_{d=0}^\infty \left(\sum_{i=1}^n a_i^{kd}\right) z^d,$$
which has a pole within the circle $|z| \leq r^{-1/k}$. (An elementary proof is also possible.)

**Fourth solution.** (Communicated by Noam Elkies.) Since $\sum_k a_k^2 = 2$, for each positive integer $k$ we have $a_k^2 \leq 2$ and so $a_k^4 \leq 2a_k^2$, with equality only for $a_k^2 \in \{0,2\}$. Thus to have $\sum_k a_k^4 = 4$, there must be a single index $k$ for which $a_k^2 = 2$, and the other $a_k$ must all equal $0$. But then $\sum_k a_k^{2m} = 2^m \neq 2m$ for any positive integer $m > 2$.

**Remark.** Manjul Bhargava points out it is easy to construct sequences of complex numbers with the desired property if we drop the condition of absolute convergence. Here is an inductive construction (of which several variants are possible). For $n = 1, 2, \ldots$ and $z \in \mathbb{C}$, define the finite sequence
$$s_{n,z} = \left(\frac{1}{z} e^{2\pi ij/n} : j = 0, \ldots, n-1\right).$$

This sequence has the property that for any positive integer $j$, the sum of the $j$-th powers of the terms of $s_{n,z}$ equals $1/z^j$ if $j$ is divisible by $n$ and $0$ otherwise. Moreover, any partial sum of $j$-th powers is bounded in absolute value by $n/|z|^j$.

The desired sequence will be constructed as follows. Suppose that we have a finite sequence which has the correct sum of $j$-th powers for $j = 1, \ldots, m$. (For instance, for $m = 1$, we may start with the singleton sequence $1$.) We may then extend it to a new sequence which has the correct sum of $j$-th powers for $j = 1, \ldots, m+1$, by appending $k$ copies of $s_{m+1,z}$ for suitable choices of a positive integer $k$ and a complex number $z$ with $|z| < m^{-2}$. This last restriction ensures that the resulting infinite sequence $a_1, a_2, \ldots$ is such that for each positive integer $m$, the series $a_1^m + a_2^m + \cdots$ is convergent (though not absolutely convergent). Its partial sums include a subsequence equal to the constant value $m$, so the sum of the series must equal $m$ as desired.