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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

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  • http://math.stackexchange.com/questions/30774/prove-that-a-n-0-for-all-n-if-sum-a-kn-0-for-all-k-geq-1 –  Jun 11 '15 at 14:45
  • does the notation of $a_{mk}$ means $a_{m_k}$? – Chilote Jun 24 '15 at 00:38
  • @Chilote : No , it does not , it means $mk$ , the product –  Jun 24 '15 at 02:01
  • Can you explain a little more the definition of $S_n(m)$? I am confused about the index. Can you show an explicit example? – Chilote Jun 24 '15 at 03:06
  • In the definition of $S_n(m)$ how does the sum depend on $n$? Is it a finite sum up to $m=n$? Shouldn't it be $S_n(k)$ because $m$ is the dummy variable in the sum? – Michael Burr Jun 25 '15 at 20:19
  • @MichaelBurr : I have explicitly written " where the sum is over the positive integers relatively prime to all positive integers not exceeding $n$ " –  Jun 26 '15 at 01:50

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