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How can one prove that if $\sum\limits_{n=1}^\infty a_n $ converges and $a_n \geq |b_n|$ only for non-prime values of $n$, $\sum\limits_{n=1}^\infty b_n $ is not necessarily convergent ?

I'm not sure how to attack this problem, but it really seems pretty interesting . Any ideas ?

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  • $\begingroup$ The “not necessarily convergent” part suggests you should find an example where $\sum_{i=1}^\infty a_n$ converges, $a_n \geq |b_n|$ if and only if $n$ is prime, and $\sum_{i=1}^\infty b_n$ diverges. $\endgroup$ – Matthew Leingang Feb 7 '15 at 21:18
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Consider $$a_{n}=\frac{1}{n^{2}},\, b_{n}=\begin{cases} \frac{1}{p}, & n=p\textrm{ is a prime}\\ \frac{1}{n^{2}}, & \textrm{otherwise} \end{cases}$$ then$$a_{n}\geq b_{n}$$ only if $n$ is non prime, but$$\underset{n\geq1}{\sum}b_{n}\geq \underset{p}{\sum}\frac{1}{p}=\infty.$$

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You should look for a counter-example. Let $\sum_{n=1}^{\infty} a_n$ be any convergent series, and take for instance $b_p=a_p$ for all prime numbers $p$.

Now, you are free to choose $b_n$ for non-prime $n$ however you want to. Can you make some choice that makes $\sum_{n=1}^{\infty} b_n$ divergent?

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  • $\begingroup$ I'm sorry, I meant non-prime values, but I think that doesn't make any difference $\endgroup$ – dean.v Feb 7 '15 at 21:32

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