Let $\sum _{n=1}^{\infty}a_n$ be an absolutely convergent series of real terms such that $\sum _{n=1}^{\infty}a_{kn}=0,\forall k \ge 1$ . For $m,n\in\mathbb N , S_n(m):=\sum a_{mk}$ , where the sum is over the positive integers relatively prime to all positive integers not exceeding $n$. How to prove that $S_n(m)=S_{n+1}(m)$ , when $n+1$ is not prime and $S_{n+1}(m)=S_n(m)-S_n((n+1)m)$ otherwise ? How to prove that $S_n(m)=0 , \forall m,n \in \mathbb N$ ? (In connection with $\sum_{n=1}^{\infty} a_n$ converges absolutely and $\sum _{n=1}^\infty a_{kn}=0 ,\forall k \ge 1 $ ; then $a_n=0 , \forall n \in \mathbb N $? )
EDIT :- It is not a duplicate , my approach is different