Does there exist $( x_n)_{n\in\mathbb{N}}\subset \mathbb{R}$ such that $\sum_{k=1}^{\infty} x_{kn} =1$ for every $n\in\mathbb{N}?$

Question

Does there exist $( x_n)_{n\in\mathbb{N}}\subset \mathbb{R}$ such that $\displaystyle\sum_{k=1}^{\infty} x_{kn} =1$ for every $n\in\mathbb{N}?$

I suspect so, but I am struggling to actually construct such a sequence. All I can say so far is that there must be infinitely many members of $( x_n)_{n\in\mathbb{N}}$ that are positive and infinitely many members of $( x_n)_{n\in\mathbb{N}}$ that are negative. Other than this, I have not made much progress.

Can someone sketch a construction or provide any hints of a disproof please?

Answer 1

A partial solution that may possibly get to a full solution with care (or of course may not, though I suspect the problem is solvable as per below)

One can construct an Abel summable sequence with the above properties (also in the Abel summability sense and even a bit more general) and maybe with care one can actually get a convergent sequence, st all the above series are convergent too by getting a sequence as below that satisfies Littlewood's Tauberian condition $n|x_n| \le C$; this, of course, implies $n|x_{nk}| \le C/k$ so all the subsequences $x_{nk}, n \ge 1, k$ fixed satisfy the Tauberian property so all the OP series converge (to $1$) if they are Abel summable (to $1$).

Let $S$ the set of all roots of unity which is countable, so there exists a non-negative integrable function on the unit circle $f \ge 0$ st if $f(re^{it})$ is its (nonnegative) harmonic extension to the unit disc (Poisson integral of $f$ or simply $f(re^{it})=\sum_{n \in \mathbb Z} a_nr^{|n|}e^{int}$ where $\sum a_ne^{int}$ is the Fourier series of $f$), then if $|w_k| \le 1, w_k \to w \in S$ we have $f(w_k) \to \infty$

This a general theorem valid for all sets of measure $0$ on the circle and is easily proven directly for $S$ by finding for every $m \ge 1$ an open circle set $G_m$ containing $S$ of circle measure at most $1/m^4$ and then define $f_m(w)=m^2, w \in G_m, f_m(w)=0$ otherwise and take $f=\sum f_m$.

Now considering $g$ the harmonic conjugate of $f$ in the unit disc and $F(z)=e^{-f-ig}$, we have that $F(0) \ne 0$ and $F(w_k) \to 0, w_k \to w \in S, |w_k|< 1$, while $F$ is analytic in the unit disc and bounded by $1$ in absolute value

Hence $G(z)=1-F(z)/F(0)$ satisfies $G(0)=0, G(w_k) \to 1, w_k \to w \in S, |w_k|< 1$ and if $G(z)= \sum_{n \ge 1} x_nz^n$ one has that if $k \ge 1$ is fixed $$\sum_{n \ge 1}x_{nk}z^{nk}=\frac{1}{k}\sum_{\omega^k=1}G(\omega z)$$

From the properties of $G$ it follows that if $z_m \to 1, |z_m| < 1$, then $$\sum_{n \ge 1}x_{nk}z_m^{nk} \to 1, m \to \infty$$ so the sequence $x_{n}$ satisfies the required properties in an Abelian sense (and a bit more as the convergence is allowed on any disc sequence going to $1$, not only radially or nontangentially)

Answer 2

We can get a sequence that works with a "just do it" sort of method, although we'll have to assume some conjectural results about the distribution of prime numbers.

The idea is that for each $k$, we'll control the sum $\sum_{n=1}^\infty x_{kn}$ by controlling only $x_{kp}$ for all prime $p>k$, and bounding the other terms enough that the rest of the sum doesn't get too out of hand for us to deal with using our prime-indexed terms of wiggle room.

More precisely, let's set $x_1=0$ (to be ignored hereafter), for all greater $n$ let $\textrm{gpf}(n)$ be the greatest prime factor of $n$, and define $f(n)=n/\textrm{gpf}(n)$. Then we'll enforce that for every $n\ge 2$:

$$|x_n|\le \ln(n)^{-3f(n)}$$

So for every prime $p$ we have $|x_p|<\frac1{\ln(p)^3}$, for every $p>2$ we have $|x_{2p}|<\frac1{\ln(p)^6}$, etc.

Subject to these constraints, we'll greedily choose each $x_n$ so as to make the partial sum for $k=f(n)$ as close to $1$ as possible. So if the partial sum $\sum_{i=1}^{\textrm{gfp}(n)-1}x_{f(n)\cdot i}=S$, then

$$x_n=\begin{cases} \ln(n)^{-3f(n)} & S<1-\ln(n)^{-3f(n)} \\ 1-S & |S-1| < \ln(n)^{-3f(n)} \\ -\ln(n)^{-3f(n)} & S>1+\ln(n)^{-3f(n)} \end{cases}$$

That's the sequence! The tricky part is showing that it has the desired property.

Pick an arbitrary $k$, and for convenience let $K_n = x_{kn}$. The sum $\sum_{n=1}^\infty K_n$ (which we want to show equals $1$) has some terms up to $K_k$ that we'll ignore except to note that they give us some finite partial sum to start off with. After that, $K_n$ is bounded in absolute value by $\ln(kn)^{-3(k+1)}$, except for $K_p$ for $p>k$, which we use to correct any problems incurred by the previous terms with a magnitude up to $\ln(kp)^{-3k}$.

We want it to be the case in the limit that:

• The divergence in the sum between prime-indexed terms is $o(1)$, so the sum doesn't oscillate.
• The prime terms are large enough to undo any drift caused by the intermediate values.
• The prime terms are sufficiently larger than said drift that we can use the leftover room to undo any drift that happened at the start of the sum, before the above limiting conditions held.

How far off can our sum get between $K_{p_n}$ and $K_{p_{n+1}}$? Let $g_n = p_{n+1}-p_n-1$ be the $n^\textrm{th}$ prime gap. In the worst case, all $g_n$ intermediate terms are pointing in the same direction, so they move the partial sum by at most

$$\sum_{i = p_n}^{p_{n+1}-1}\ln(ki)^{-3(k+1)} < g\cdot \ln(kp_n)^{-3(k+1)}$$

We'd like for $\ln(kp_{n+1})^{-3k} > 2g\cdot \ln(kp_n)^{-3(k+1)}$, since then the prime terms will be able to undo any inter-prime drift and have enough magnitude left over to pull the sum back by at least $0.5\ln(kp_{n+1})^{-3k}$, which goes to infinity when added over all primes.

This means that it would suffice to have, for sufficiently large $n$,

$$g < \ln(p_n)^3 \le \ln(kp_n)^{3} = \frac{\ln(kp_n)^{3(k+1)} }{\ln(kp_n)^{3k}} < \frac{2\ln(kp_n)^{3(k+1)}}{\ln(kp_{n+1})^{3k} } =\frac{2\ln(kp_{n+1})^{-3k} }{\ln(kp_n)^{-3(k+1)}}$$

where the last inequality is due to the fact that $\frac{\ln(kp_{n+1})}{\ln(kp_n)}$ approaches $1$ for large $p$ and hence eventually drops below $2^{1/3k}$.

Such a strong bound on prime gaps is well beyond the reach of modern number theory, even assuming RH. However, it is implied by Cramer's conjecture; my impression is that while Cramer's conjecture may be somewhat too strong it is believed that prime gaps are still $o(\log(p)^c)$ for every $c>2$, and for our purposes we need only $c=3$ (or indeed any finite $c$ we like by changing our enforcement criterion above).

Note that in all likelihood one doesn't need anywhere near this strong of a bound, because the inter-prime terms will have effectively random signs and the sum will drift away much much slower. But I don't see a way to guarantee this variation, so we need a very powerful hammer to make sure things work out.