Convergence of the series involving Mobius functions $\sum_{k,d} \mu(d) x_{kd}$

Question

(I originally asked this question here, but the problem appears much more difficult than I think after a moment of thought, so I think it might be more suitable to post it here. Please tell me if this is not the right place to ask.)

Let $\sum x_n$ be an absolutely convergence sequence, then obviously, from the elementary properties of Mobius function $\mu$ that $\sum_{d|n} \mu (d)=0$ for $n\neq 1$, $$ \sum_{n=1}^\infty \sum_{d|n} \mu(d)x_n =x_1. $$ However, if we alter the order of summation, i.e. $$ \sum_{d=1}^\infty \sum_{k=1}^\infty \mu(d) x_{kd}, $$ where $kd$ is the product of $k$ and $d$, will we get the same sum, or does the sum diverge?

One way of proving that we can exchange the order of summation is that $$ \left|\sum_{kd\geq M, k\leq K,d\leq D} \mu(d) x_{kd}\right| <\epsilon $$ for all $K,D$ and sufficiently large $M$. But then $\mu$ does not cancel out as nicely, and the bound on the sum of $\mu$'s using Merten's function is too weak.

Since the proof seems difficult, I attempt to find a divergent counterexample. I try to use the fact proven in an answer below this MO question that the Dirichlet series of Mobius function $\mu$, $$ \sum_{n=1}^\infty \mu(n)n^{-s}, $$ diverges for $0<\Re s <1/2$.

Define the sequence $b_n$ by $$ b_n=n^{-1/4}-(n+1)^{-1/4}. $$ Clearly $\sum_{n=r}^\infty b_n = r^{-1/4}$.

Now define $a_{k}$ as follows.

1. $a_1=b_1$ .
2. Suppose that $a_1,\ldots, a_{l}$ are already defined, such that $a_l=b_n$ for some $n$. Then let $l'>l$ be a multiple of $(n+1)!$, and $a_{l'}:=b_{n+1}$. Furthermore, let $a_{k}=0$ for $k$ strictly between $l$ and $l'$.

This recursive definition gives a sequence $a_n$, such that the sequence of sums $S_n=\sum_{k=1}^{\infty} a_{nk}$ decreases slower than $n^{-1/4}$ (again $nk$ denotes the product of $n$ and $k$). However, this does not make $\sum_{d=1}^\infty \sum_{k=1}^\infty \mu(d) x_{kd}$ a Dirichlet series, so even if the decrease in magnitude is slow, we cannot conclude that it is divergent. Positive and negative terms might be cancelled out.

So, both proof and counterexample are difficult - the bounds and constructions are always a little bit worse then desired.

Are there any known results about the convergence of the sum $\sum_{d=1}^\infty \sum_{k=1}^\infty \mu(d) x_{kd}$?

Answer 1

This question was considered by Wintner (1945). On pages 16-18 of the linked document, he observes that $\sum_n 2^{\omega(n)}|x_n|<\infty$ implies the absolute convergence of the second display (here $\omega(n)=\sum_{p\mid n}1$), hence in this case the second display converges to $x_1$. Moreover, based on Toeplitz's norm principle, he constructs an example with $\sum_n |x_n|<\infty$ for which the OP's second display diverges.