12
$\begingroup$

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.

$\endgroup$
8
  • 2
    $\begingroup$ I wonder if it is possible that $\sum_{i = 1}^{\infty} a_i^k$ is conditionally convergent for all $k$. $\endgroup$
    – Martin R
    Feb 18, 2023 at 21:13
  • $\begingroup$ 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 $k > 1$. $\endgroup$ Feb 18, 2023 at 21:17
  • 2
    $\begingroup$ 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). $\endgroup$
    – Martin R
    Feb 18, 2023 at 21:59
  • 4
    $\begingroup$ 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. $\endgroup$ Feb 19, 2023 at 17:31
  • 1
    $\begingroup$ 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! $\endgroup$ Feb 21, 2023 at 18:27

2 Answers 2

6
$\begingroup$

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 $k$

$\endgroup$
5
$\begingroup$

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 $$\sum_{d|n, d<n}f(d)r_d^n+f(n)r_n^n=0$$ 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 $$|r_n|^n \le \frac{\sum_{d|n, d<n}f(d)|r_d^n|}{f(n)} \le \frac{\sum_{m<n}f(m)}{f(n)} \le \frac{1}{n^{2n}}$$

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 $n<N$ 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 $$\sum_{d|k, d<k}f(d)r_d^k+f(k)r_k^k=0$$

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

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .