Let $\sum x_n$ be a power series which behaves sufficiently nicely, for example, absolutely convergent. Can we deduce that the double series $$ \sum_{d|n} \mu(d) x_{n} $$ converges in the sense described below to zero? From the property of Mobius function $\mu$ that $\sum_{d|n} \mu (d)=0$, the only possible limit of this series is zero. But to show that it converges is a bit difficult, since $\mu$ does not necessarily sum nicely to zero in partial sums.
Definition of convergence: we say that a double series $\sum a_{ij}$ converges to $l$ if for all $\epsilon>0$ there exists $N>0$ such that for all finite set $I\subseteq \mathbb N \times \mathbb N$ satisfying $\{0,1,\ldots, N\}^2 \subseteq I$, we have $$ \left| \sum_{(i,j)\in I} a_{ij} - l \right|<\epsilon. $$
In particular, for this question, we have $$ a_{ij}=\mu(i) x_j \text{ if }i|j,\\ a_{ij}=0 \text{ otherwise. } $$
Alternatively we can consider the series $$ \sum_{d, m} \mu(d) x_{md}, $$ where $md$ is the product of $m$ and $d$.
$$\sum_{n\ge 1}\sum_{d|n} \mu(d) x_{n}= x_1$$ Did you mean $$\sum_{d\ge 1}\mu(d)\sum_{m\ge 1} x_{md}$$ ? If $x_n=O( n^{-r})$ for some $r> 1$, since $\tau(n)=O( n^{(r-1)/2})$ then $$\sum_{n\ge 1}\sum_{d|n} |\mu(d) x_{n}|\le \sum_{n\ge 1}|x_{n}|\tau(n)$$ converges so we can change the order of summation to get that both series are equal.
The double series $\sum_{n,d} \mu(d) x_n$ converges independently of the order of summation (thus to $x_1$) iff it converges absolutely iff $\sum_n |x_n| 2^{\omega(n)}$ converges.
With $$\cases{x_n = k^{-1/4}-(k+1)^{-1/4} \ if \ n= \prod_{p\le k} p\\x_n=0 \ otherwise}$$ and $Lpf$ the largest prime factor we get that $$\sum_{d\ge 1}\mu(d)\sum_{m\ge 1} x_{md}=\sum_{d\ge 1} \mu(d)Lpf(d)^{-1/4}$$ which diverges (proof ?)