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Let us consider two sequences of real numbers $a_n$ and $b_n$, about which we only know that: $$\sum_{1}^{\infty}a_n = 0$$ and that all $b_n > 0$, with $b_{n+1} > b_n$. Can it be proved that there cannot exist a $b_n$ sequence with said features, such that also $$\sum_{1}^{\infty}a_n b_n = 0 \;\;?$$ or is such hypothesis unjustified, and there instead exist counterexamples ?

Thanks

Following Mark's reply, I am reformulating into a somewhat different question, which is also of interest to me.

Given the above conditions for $a_n$ and $b_n$, would it be possible to find additional general conditions for the $a_n$'s which will then be sufficient to ensure that $$\sum_{1}^{\infty}a_n b_n \neq 0 \;\;?$$

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Consider the series $a=1-1+1/4-1/4+1/9-1/9...+1/n^2-1/n^2...$ and $b=1+1+2+2+3+3+4+4...+n+n...$ . Their "product" is $1-1+1/2-1/2....1/n-1/n...$ converges to 0.If you really want $b_{n+1}\gt b_n$, then instead $...n+n...$ consider the series $...n+(n+1/n^2)...$.

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  • $\begingroup$ but doesn't your $a_n$ in fact resemble a bit too closely 0+0+0+0+.... ? thanks anyway, as you made me realize that I shall perhaps better reformulate my question with more constraints on ${a_n}$ $\endgroup$
    – Luca
    Commented Mar 15, 2020 at 12:53
  • $\begingroup$ No it does not. If you want all $a_n$ to be dfferent in absolute value, add any other series where the sum is 0 and the terms converge to 0 faster than $1/n^4$. $\endgroup$
    – user6976
    Commented Mar 15, 2020 at 13:05

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