Is there an infinite sequence of complex numbers $a_1, a_2, a_3, \ldots$ such that $a_1^m + a_2^m + a_3^m + \ldots = m$ for every positive integer m?

Question

A similar question was asked for the case where all $a_i$ are real here: Is there an infinite sequence of real numbers $a_1, a_2, a_3,... $ such that ${a_1}^m+{a_2}^m+a_3^m+...=m$ for every positive integer $m$?

Answer 1

This was problem $B$-$1$ on the Putnam exam in $2010$. You can find the detailed solution here. In essence, the answer is that there is no such sequence of $a_i$ exists if we assme that the series $\sum a_i^m$ converges absolutely, and the solution manual provides a proof. It also provides a construction of a particular series of $a_n$ which satisfy the conditions, despite $\sum a_i^m$ not being absolutely convergent.

Third solution. We generalize the second solution to show that for any positive integer $k$, it is impossible for a sequence $a_1, a_2,\dots$ 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 $|a_1|,\dots,|a_n| \leq 1$, 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 $r = \max\{|a_1|,\dots,|a_n|\} > 1$, we obtain another contradiction because for any $\epsilon > 0$, $$ \limsup_{d \to \infty} (r-\epsilon)^{-kd} \left| \sum_{i=1}^n a_i^{kd} \right| > 0. $$ 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.)

Answer 2

Assume that $(a_k)$ is a sequence of complex numbers such that

• $\sum_{k=1}^{\infty} a_k^m$ converges for each $m$, and

• there exists $m_0 \geq 1$ such that $\sum_{k=1}^{\infty} |a_k|^{m_0} < \infty$.

Then it is impossible to have $\sum_{k=1}^{\infty} a_k^m = m$ for all $m \geq 1$ (or even for all sufficiently large $m$). To this end, we assume otherwise and show that this leads to a contradiction.

Let $R = (\sup_{k\geq 1} |a_k|)^{-1} $. Then for any $|z| < R$,

$$ \frac{z}{(1-z)^2} - \Biggl( \sum_{m=1}^{m_0 - 1} m z^m \Biggr) = \sum_{m=m_0}^{\infty} m z^m = \sum_{m=m_0}^{\infty} \sum_{k=1}^{\infty} a_k^m z^m = \sum_{k=1}^{\infty} \frac{(a_k z)^{m_0}}{1 - a_k z}. $$

In the second step, we utilized the Fubini's theorem. The right-hand side converges locally uniformly on any compact subset of $\mathbb{C} \setminus A$ with $A = \{a_k^{-1} : k \geq 1 \text{ and } a_k \neq 0 \}$. Then for any bounded, simply connected region $\mathcal{D}$ such that $1 \in \mathcal{D}$ and $\partial\mathcal{D}$ avoids $A$,

\begin{align*} 1 = \frac{i}{2\pi} \int_{\partial\mathcal{D}} \Biggl( \sum_{m=m_0}^{\infty} m z^m \Biggr) \, \mathrm{d}z = \frac{i}{2\pi} \sum_{k=1}^{\infty} \int_{\partial\mathcal{D}} \frac{(a_k z)^{m_0}}{1 - a_k z} \, \mathrm{d}z = \sum_{k \, : \, a_k^{-1} \in \mathcal{D}} \frac{1}{a_k}. \end{align*}

Since $A$ is discrete, this implies that $A \subseteq \{1\}$ and there is exactly one non-zero element in $(a_k)$, which takes value $1$. Then it is obvious that such sequence cannot satisfy $\sum_{k=1}^{\infty} a_k^m = m$ for large $m$. $\square$

Without absolute convergence condition, I even have no idea whether the statement is true or not.