If $\sum\limits_nx_{kn}=0$ for all $k$ and $\sum\limits_n|x_n|$ converges then $x_n=0$ for all $n$, but what if $\sum\limits_n|x_n|$ diverges?

Question

I am trying to answer the following question, specifically, the second part:

Let $(x_{n})^{\infty}_{n=1}$ be real sequence such that $\sum^{\infty}_{n=1}|x_{n}|$ converges and, for each $k\in\mathbb{N}$, $\sum^{\infty}_{n=1}x_{kn} = 0$. Show that $x_{n} = 0$ for all $n$.

What if we no longer require $\sum^{\infty}_{n=1}|x_{n}|$ to converge?

Source: https://www.dpmms.cam.ac.uk/study/IA/Numbers+Sets/2015-2016/examples-NS-15-3.pdf

Solution to first part:

Let $p_{i,k}$ denote the $i^{th}$ prime grater than $k$ (e.g $p_{1,7} = p_{1,8} = p_{2,6} = 11$).

Define $s_{j,k} = \sum_{n\in k\mathbb{N} \setminus A_{j,k}} x_{n}$ where $A_{j,k} = \{n \in k\mathbb{N} | \exists i \leq j\,\,\,\, s.t. \,\, n|p_{i,k} \}$.

$s_{j,k} = \sum_{n\in k\mathbb{N} \setminus A_{j,k}} x_{n} = \sum_{n\in A_{j,k}}x_{n}$ because $\sum x_{kn} = 0$.

Using Inclusion-Exclusion:

$s_{j,k} = \sum^{m}_{r=1}\bigg((-1)^{r-1}\sum_{I \subset \{p_{1,k}, p_{2,k}, ..., p_{j,k}\}, |I|=r}\big(\sum_{n\in A_{i,k},i \in I}x_{kn}\big)\bigg)$.

We see the innermost sum is $0$. Hence $s_{j,k} = 0$.

$s_{i,k} = x_{k} + \sum_{n \in B_{k}}x_{n} = 0$ where $B_{k} \subset \{p_{j+1, k}, p_{j+1,k}+ 1,...\}$.

Hence:

$|x_{k}| = |\sum_{n\in B_{k}} x_{n}| \leq \sum_{n \in B_{k}} |x_{n}| \leq \sum_{n \geq p_{j+1,k}}|x_{n}|$

By taking limits as $j \to\infty$ we have $|x_{k}| = 0$. So $x_{n} = 0 \, \, \, \forall n$.

Unsure on second part of question.

Answer 1

If we do not require absolute convergence, then there are nonzero sequences $(x_n)$ such that

$$\sum_{n = 1}^{\infty} x_{kn} = 0$$

for all $k \in \mathbb{N}$ (assuming that here $0 \notin \mathbb{N}$; but if it is, things still work if we have $x_0 = 0$).

One example is provided by the Liouville function $\lambda$. For a positive integer $n$, let $\Omega(n)$ be the number of prime divisors of $n$, counting multiplicity. So for example $\Omega(4) = 2 = \Omega(10)$ and $\Omega(30) = 3 = \Omega(12)$. Then $\lambda(n) := (-1)^{\Omega(n)}$. From the identity

$$\frac{\zeta(2s)}{\zeta(s)} = \prod_p \frac{1 - p^{-s}}{1 - p^{-2s}} = \prod_p \frac{1}{1 + p^{-s}} = \sum_{n = 1}^{\infty} \frac{\lambda(n)}{n^s}$$

for $\operatorname{Re} s > 1$ and the fact that the Riemann $\zeta$ function has a pole at $1$ and is holomorphic at $2$, it follows that if the series

$$\sum_{n = 1}^{\infty} \frac{\lambda(n)}{n} \tag{1}$$

converges, the sum of the series is $0$. The fact that this series is convergent is nontrivial, it is equivalent - in the sense that each can be relatively easily derived from the other, significantly more easily than proving either - to the prime number theorem. Since the Liouville function is completely multiplicative, we have

$$\sum_{n = 1}^{\infty} \frac{\lambda(kn)}{kn} = \sum_{n = 1}^{\infty} \frac{\lambda(k)}{k}\cdot \frac{\lambda(n)}{n} = \frac{\lambda(k)}{k}\sum_{n = 1}^{\infty} \frac{\lambda(n)}{n} = 0$$

for all $k \in \mathbb{N}$.

A related example is given by the Möbius function, i.e. $\mu(n) = 0$ if $n$ is divisible by the square of a prime, and $\mu(n) = (-1)^r$ if $n$ is the product of $r$ distinct primes. Define

$$x_n := \frac{\mu(n)}{n}.$$

In 1897, von Mangoldt proved that $\sum_{n = 1}^{\infty} x_n$ converges (to $0$). Again this is equivalent - in the same sense as above - to the prime number theorem. Landau proved 1905 that all series

$$\sum_{n = 1}^{\infty} x_{k n + \ell}$$

with $k \in \mathbb{N}$ and $0 \leqslant \ell < k$ are convergent. It remains to see that

$$S_k := \sum_{n = 1}^{\infty} x_{kn} = 0 \tag{$\ast$}$$

for all $k \in \mathbb{N}$. We show that by induction over the number of prime factors of $k$.

The base case $k = 1$ is von Mangoldt's theorem.

In the induction step, we assume that $(\ast)$ holds for a given $k$ and show that it also holds for $k' = p\cdot k$, where $p$ is an arbitrary prime. If $p$ divides $k$, the result is easy, for then $\mu(pkn) = 0$ for all $n$, since $p^2$ divides $pk$ and hence $pkn$. If $p$ doesn't divide $k$, we compute

\begin{align} S_{pk} &= \sum_{n = 1}^{\infty} \frac{\mu(pkn)}{pkn} \\ &= \sum_{n\notin p\mathbb{N}} \frac{\mu(pkn)}{pkn} \tag{$p\mid n \Rightarrow \mu(pkn) = 0$} \\ &= -\frac{1}{p} \sum_{n\notin p\mathbb{N}} \frac{\mu(kn)}{kn} \tag{$p\nmid kn \Rightarrow \mu(pkn) = -\mu(kn)$} \\ &= \frac{1}{p} \sum_{n \in p\mathbb{N}} \frac{\mu(kn)}{kn} \tag{$S_k = 0$} \\ &= \frac{1}{p} \sum_{m = 1}^{\infty} \frac{\mu(kpm)}{kpm} \tag{$n = pm$} \\ &= \frac{1}{p} S_{pk}, \end{align}

from which we conclude $S_{pk} = 0$.