# 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?

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 $m$?

• there is a comment under the accepted answer giving a sketch of applying the argument to the complexes – Will Jagy Oct 13 '19 at 2:10
• @WillJagy Yeah, I saw that comment, but the argument seems false in the case where there are multiple complex numbers with the same (maximal) magnitude – josh Oct 13 '19 at 2:22
• I think there is a lot of detail hidden in the comment "their sum [the multiple complex numbers with same max magnitude] acts exponentially for some [sub]sequence of n". I.e. their "angles" align frequently enough to have a subsequence grow exponential. I personally don't know how to show that. – antkam Oct 13 '19 at 14:11
• @antkam Yes, that is probably the case, but would seem to require some number theoretic machinery to show – josh Oct 13 '19 at 22:11

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.)

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.