# Does there exist an infinite sequence of complex numbers $\{a_i\}$, all of whose powers sum to zero?

Does there exist an infinite sequence $$a_1, a_2, \dots \in \mathbb{C}$$ such that for all integers $$k \ge 1$$, we have $$\sum_{i = 1}^{\infty} a_i^k = 0?$$

The statement is true if the $$a's$$ are absolutely convergent, but this question is about if we relax absolute convergence and only have conditional.

If we instead take a finite sequence and sufficiently many powers are zero, then indeed they are all zero, but this method can't work in the infinite case. [If you truncate the series at $$a_1\dots a_n$$ and only have $$n$$ equations, there is no guarantee the RHS will be small, and even if there was, there is no guarantee that the coefficients $$e_i$$ of the polynomial will be small, and even if there was, a polynomial with small coefficients can still have large roots that do not approach zero (eg. $$x^n - 1/2^n$$ has roots $$x = 1/2$$.)]

In the style of this excellent video, you could ask if the set of vectors $$v_2 = (a_2^1, a_2^2, \dots), v_3 = (a_3^1, a_3^2, \dots)$$, each of which is in $$l_2$$, can be linearly combined to produce $$v_1 = (-a_1^1, -a_1^2, \dots)$$. Per the video the answer is indeed yes! - but this does not guarantee that the linear combination will be of equal weights $$v_1 = 1\cdot v_2 + 1\cdot v_3 + \dots$$. [Actually, an equal weight sum of vectors in $$l_2$$ is not necessarily even in $$l_2$$ itself - eg., if $$v_i$$ is the vector of all zeros except for position $$i$$ which is 1, then $$v$$ is a vector of all 1s, which is not in $$l_2$$.]

To try to use complex analysis, it's pretty clear that for any analytic (on the unit disc?) function $$f(z) = \sum_{j = 1}^{\infty} c_j z^j$$ for which $$f(0) = 0$$, then $$\sum_i^{\infty} f(a_i) = 0$$. Or, if you want to remove the $$f(0) = 0$$ condition, for all analytic functions $$g$$, we have $$\sum_{i=0}^{\infty}a_i g(a_i) = 0$$. This works for ANY g! Surely at this point it should be obvious that the $$a_i$$s must be zero, but alas I can't see it.

The same would hold if $$a_i$$ was replaced with $$\overline{a_i}$$, so that $$\sum_{i=0}^{\infty}\overline{a_i} g(\overline{a_i}) = 0$$. But I don't think this gets us anywhere, since we already know that $$a_i$$ converges to zero, so we can't use some trick to show that for some $$s$$, then $$g(s) = 0$$ for all $$s$$, contradiction, so all $$a_i$$ are zero. And we have a fixed set of $$a_i$$, so we can't move them around to such an $$s$$ and obtain that $$g(s) = 0$$.

And we have to use analytic functions - the Unit Disc Stone Weierstrass theorem says that $$f(z)$$ can be approximated by a polynomial $$p(z, \overline{z})$$, a polynomial in two variables $$z, \overline{z}$$. This makes things hard - because we don't have any information on the magnitudes of $$a_i$$ other than that (WLOG) they are less than 1.

Actually, if the $$a_i$$s were real numbers (and we eschew the trivial inequality to finish our proof immediately, haha), the usual Stone Weierstrass would almost give a slick proof, but not quite. If we use the same polynomial $$p$$ to approximate an arbitrary continuous function $$f$$, so that $$|p(x)-f(x)| \le \epsilon$$ for all $$x$$, we'd have the same absolute error bound but an infinite number of terms, so the infinite sum $$0 = \sum_{i=0}^{\infty} p(a_i)$$ would not be close to $$\sum_{i=0}^{\infty} f(a_i)$$ as is true in the finite case. If we ignore this issue anyway and assume that $$\sum_{i=0}^{\infty} f(a_i)$$ is small, then we can construct a neighborhood $$[-\epsilon, \epsilon]$$ around zero which contains all but a finite nonzero number of the $$a_i$$ at the beginning of the sequence; then we can construct a continuous function which is zero on this neighborhood but equal to one everywhere else, and we'd obtain a contradiction (the sum would be strictly positive and more than one, vs close to 0), showing that no such interval exists, and thus all the $$a_i$$ have to be equal, and then it follows that $$a_i = 0$$ for all $$i$$.

I wonder if there is a nonzero solution where the $$a_i$$ are only conditionally convergent but not absolutely. On the other hand, from the vectors idea above, I think a nonzero solution is impossible because we have a uniform summation of the powers of $$a_i$$, which would prevent the resulting sum from lying in $$l_2$$ like $$v_1 = (-a_1^1, -a_1^2, \dots)$$ does.

• I wonder if it is possible that $\sum_{i = 1}^{\infty} a_i^k$ is conditionally convergent for all $k$. Feb 18, 2023 at 21:13
• Per the first link, it seems that any non-zero solution would fail to be absolutely convergent at $k=1$ but i don't know how if it implies for $k > 1$. Feb 18, 2023 at 21:17
• It is possible: If $a_n = \exp(i n \pi \alpha)/\ln(n)$ with real but irrational $\alpha$ then $\sum_{n=1}^\infty a_n^k$ is convergent (by Dirichlet's test), but not absolutely convergent. Which means that your question can not be reduced to the case of absolutely convergent series (at least not obviously). Feb 18, 2023 at 21:59
• See Andrew Lenard, A nonzero complex sequence with vanishing power-sums, Proceedings of the American Mathematical Society 108 (1990), 951–953; as well as the generalizations in William M. Priestley, Complex Sequences Whose “Moments” all Vanish, Proceedings of the American Mathematical Society 116 (1992), 437–444. Feb 19, 2023 at 17:31
• Wow, crud, I was totally ready for the answer to be "no" but without a constructive proof. That a constructive proof, that a high schooler could understand, exists, is incredible. Thank you all for your help and interest! Feb 21, 2023 at 18:27

As ho boon suan said in a comment, an example of such a sequence has been given in

They construct a sequence of non-zero complex numbers $$(a_n)$$ such that $$\sum_{n=1}^\infty a_n^k = 0$$ for all positive integers $$k$$.

Here is a sketch of that construction. First, finite sequences $$s_0, s_1, s_2, \ldots$$ are recursively defined as follows

• $$s_0 = 1$$.

• $$s_1$$ is $$s_0$$, followed by a copy of $$s_0$$ multiplied with $$\alpha_1 = \exp(i\pi) = -1$$: $$s_1 = 1, -1 \, .$$

• $$s_2$$ is $$s_1$$, followed by $$2^2 = 4$$ copies of $$s_1$$ where each term is multiplied with $$\alpha_2 = \exp(i\pi/2)/2 = i/2$$: $$s_2 = 1, -1, i/2, -i/2, i/2, -i/2, i/2, -i/2, i/2, -i/2 \, .$$

• $$s_3$$ is $$s_2$$, followed by $$3^3 = 27$$ copies of $$s_2$$ where each term is multiplied with $$\alpha_3 = \exp(i\pi/3)/3$$, that are $$10 \cdot (1+27) = 280$$ terms, starting with $$s_3 = 1, -1, i/2, -i/2, i/2, -i/2, i/2, -i/2, i/2, -i/2 \\ \exp(i\pi/3)/3, -\exp(i\pi/3)/3, \exp(i\pi/3)/3 \cdot i/2, \ldots$$

• ...

• $$s_j$$ is $$s_{j-1}$$, followed by $$j^j$$ copies of $$s_{j-1}$$ where each term is multiplied with $$\alpha_j = \exp(i\pi/j)/j$$.

The construction is such that

• The sum of all terms in $$s_1$$ vanishes, and the same is true for the sum of all terms in $$s_j$$, $$j \ge 1$$.
• The sum of the squares of all terms in $$s_2$$ vanishes, and the same is true for the sum of squares of all terms in $$s_j$$ with $$j \ge 2$$.
• ...
• Generally, the sum of the $$k$$-th powers of all terms in $$s_j$$ vanishes for $$j \ge k$$.

Each of these sequences is an extension of the previous ones, so that one can define $$a_n$$ as the $$n$$-th term in $$s_j$$ for sufficiently large $$j$$.

For each $$k \ge 1$$, the sequence $$(a_n)$$ can be viewed as the concatenation of infinitely many copies of $$s_k$$, scaled with complex factors of decreasing modulus. It follows that $$\sum_{n=1}^\infty a_n^k = 0$$ for all positive integers $$k$$

We will construct such a sequence in three steps.

First let's have $$f: \mathbb N^* \to \mathbb N^*, f(1)=1$$ st $$n|f(n)$$ and that grows so fast so $$\sum_{m=1}^{n-1}f(m) \le f(n)/n^{2n}$$, so in particular $$f(2) \ge 16$$. Then we pick some $$|r_1| \le 1, r_1 \ne 0$$ arbitrary and inductively we construct $$r_n \in \mathbb C$$ st $$\sum_{d|n, d and we claim that $$|r_n| \le 1/n^2$$ since $$|r_2|^2=|r_1|^2f(1)/f(2) \le 1/16$$ so $$|r_2| \le 1/4$$ and then inductively we have $$|r_n|^n \le \frac{\sum_{d|n, d

We note that if $$\omega_{n1}=1,..\omega_{nn}$$ are the roots of unity of order $$n \ge 2$$, then $$\sum_{m=1}^n\omega_{nm}^k=0$$ if $$k \ne an$$ and $$\sum_{m=1}^n\omega_{nm}^{an}=n$$. Also we notice that any partial sums of the $$\omega^k$$'s is at most $$n$$ in absolute value, since they are roots of unity.

Now for every $$n \ge 1$$ we define the $$a$$ in blocks of $$f(n)$$ where each consecutive block of $$n$$ consists of the roots of order $$n$$ multiplied by $$r_n$$, and we repeat this precisely $$f(n)/n$$ times; so $$a_1=r_1, a_{21}=r_2, a_{22}=-r_2,a_{23}=r_2, a_{24}=-r_2,..a_{2(15)}=r_1, a_{2(16)}=-r_2$$ etc if say $$f(2)=16$$ which is lowest such allowable (and of course note that any $$r_p \ne 0$$ for prime $$p$$ so $$a_n \ne 0$$ for infinitely many $$n$$)

So in general $$a_{nl}=\omega_{nl}r_l$$ where $$l=1,..f(n)$$ and the roots of unity are extended by periodcity so $$\omega_{nl}=\omega_{n(l+n)}$$ when $$l >n$$.

Then we put together the $$a_{nl}, l=1,..f(n)$$ in one sequence in the order of $$n$$ and then of $$l$$ so we have $$a_1, a_{21}, a_{22}, a_{23},..a_{28}, a_{31},...a_{3f(3)},...$$ and we claim that $$\sum_{n \ge 1}a_n=a_1, \sum_{n \ge 1} a_n^k=0, k \ge 2$$

Then we choose another sequence $$b_n$$ as above where $$b_1=-a_1$$ and it will follow that the sequence $$c_n$$ where we alternate $$c_{2n-1}=a_n, c_{2n}=b_n$$ will satisfy $$\sum c_n^k=0$$ for all $$k \ge 1$$

Note first that by construction $$\sum_{l=1}^{f(n)}a_{nl}^k=0, k \ne qn$$ and $$\sum_{l=1}^{f(n)}a_{nl}^{qn}=f(n)r_n^{qn}$$

So first let $$k=1$$ and look at the partial sums $$s_N=a_1+\sum_{2 \le m \le N}a_m$$. Since for all $$n \ge 2$$ we have that $$1$$ is not a multiple of $$n$$, any complete block of $$f(n)$$ of roots of unity of order $$n$$ vanishes so $$s_N \to a_1$$ if it converges. But any partial sums vanishes on all complete blocks so we just in general remain with the sum of the $$a_m$$ within the last block consisting of roots of unity of order $$n that may not be complete, but even there any consecutive $$n$$ terms vanish so actually we remain with a sum of at most $$n-1$$ terms which is at most $$n|r_n| \le 1/n$$ in absolute value and since $$N \to \infty$$ means $$n \to \infty$$ as $$f(n)$$ while very large is still finite, we get that $$s_n \to a_1$$

Now let's have $$k \ge 2$$ and note that again for $$n >k$$ all the blocks of roots of unity of order $$n$$ vanish, so we have that if $$s_{Nk}=\sum_{1\le m \le N}a_m^k$$ converges it does so to $$S_k$$ which is the sum of the terms up to the end of the block $$f(k)$$ of roots of unity of order $$k$$ (multiplied by $$r_k^k$$). But by construction, this sum is precisely $$\sum_{d|k, d

And now for the convergence of $$s_{Nk}$$ again we need to look only at the terms for which the blocks are of order $$n>k$$ and any complete such vanishes, while any incomplete such again has at most $$n-1$$ terms since any block of $$n$$ consecutive terms vanishes too; so any such partial sums $$\sum_{N_1 \le m \le M_1}a_m^k, N_1>f(1)+..f(k)$$ is at most $$n|r_n|^k$$ in absolute value so goes to zero with $$n \to \infty$$ since $$|r_n| \le 1/n^2$$. This shows that $$\sum a_n^k \to S_k=0$$ and we are done since $$n \to \infty$$ as $$N_1, M_1 \to \infty$$.

Choosing $$b_n$$ with $$b_1=-a_1$$ and mixing them, clearly doesn't alter the argument since now for any partial sum we can split it into the partial sums on the $$a_n$$ and the $$b_n$$'s respectively and apply the arguments above to each with of course the appropriate changes.

Hence there are indeed $$a_m \ne 0$$ st $$\sum a_m^k=0$$ for all $$k \ge 1$$ while clearly $$\sum |a_m|= \sum f(n)|r_n|$$ is wildly divergent