Prove that $a_{n}=0$ for all $n$, if $\sum a_{kn}=0$ for all $k\geq 1$ Let $\sum a_{n}$ be an absolutely convergent series such that $$\sum a_{kn}=0$$ for all $k\geq 1$.   Help me prove that $a_{n}=0$ for all $n$.
Thank you!

Let $A(p_1, \dots, p_m)$ be the set of all integers that are divisible by the first $m$ primes $p_1, \dots, p_m$.
I claim that 
$$S(m) = \sum_{n \in N \setminus A(p_1, \dots, p_m)} x_n = 0.$$
If this is true, then 
$$S(m) = x_1 + \sum_{n \in A} x_n = 0$$
where $A \subseteq \{p_{m+1}, p_{m+1}+1, \dots\}$.
That is,
$$|x_1| = \left| \sum_{x \in A} x_n \right| \leq \sum_{n \geq p_m}|x_n|.$$
Because $\sum |x_n|$ converges, if we let $m \to \infty$, we get $x_1=0$.
Next we realize that the sequence $x_{kn}$ satisfies the same conditions as $x_n$. Hence the same reasoning can be applied as above to conclude that $x_{1 \cdot k} = 0$. From this we conclude the entire sequence is zero identically.

It remains to show
$$S(m) = \sum_{n \in N \setminus A(p_1, \dots, p_m)} x_n = 0.$$
Because $\sum_n x_n = 0$, this is equivalent to showing
$$S(m) = \sum_{n \in A(p_1, \dots, p_m)} x_n = 0.$$
Using inclusion exclusion formula,
$$S(m) = \sum_{i=1}^m \left( (-1)^{i-1} \sum_{I \subseteq \{p_1, \dots, p_m\}, |I|=k} \sum_{n \in A(I)} x_n \right)$$
Where the innermost sum is over all numbers that are multiples of a subset of the primes in $\{p_1, \dots, p_m\}$. But this sum is of the form $\sum_{n} x_{nk}$ for $k$ the product of the primes in the subset, by assumption our sum is zero. Hence the entire expression for $S(m)$ is zero. This proves our claim, which then is used to prove our initial claim.
