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I have been wondering about this question for some time now and would greatly appreciate if anyone had any leads on how to proceed.

Let $\{A_{i}\}_{i \in N}$ such that for, $i \neq j$, $A_{i} \cap A_{j} = \emptyset$ and $\bigcup{A_{i}} = N$. Let $(a_{i})_{i \in N}$ be a sequence in $R$. If, for all $A_{i}$, $\sum_{j \in A_{i}}{a_{j}} \geq 0$ (taking the usual ordering in $A_{i}$), is it true that $\sum_{i \leq n}{a_{i}} \geq 0$ infinitely often?

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Let $a_0=-1$, $a_1=-1$, $a_2=1$, $a_3=-1$, $a_4=1$, $a_5=-1$, $a_6=1$ and so on. Then all partial sums are negative, alternating between $-1$ and $-2$.

Let $A_0=\{0,2\}$, $A_1=\{1,4\}$, $A_2=\{3,6\}$, $A_3=\{5,8\}$, and so on. Then for any $j$ the sum of the elements of $A_j$ is $0$.

Thus there are quite simple partitions of the natural numbers for which the desired result fails.

Remark: We can produce a convergent series that converges to $0$ with the same property. Let $a_0=-1$. Let $a_1=1/2$, $a_2=1/3$. Adding $1/4$ would make the partial sum $\ge 0$, so $a_3=-1/4$. Now we can go forward again, twice. Let $a_4=1/5$ and $a_5=1/6$. Adding $1/7$ would make the partial sum positive, so $a_6=-1/7$. Continue. Again, we can partition the natural numbers into (finite) parts so that the sum over any part is $\ge 0$.

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  • $\begingroup$ Might I ask if the result is still trivial if $\sum_{i \in N}{a_{i}}$ converges? Thanks! $\endgroup$
    – madprob
    Jul 23, 2012 at 21:43
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    $\begingroup$ @madprob: sorry, missed your message until now. Start at $-1$. Next is $1/2$, $1/3$. Adding $1/4$ would make us positive. So next is $-1/4$. Then next are $1/5$ and $1/6$. Next would put us over, so next is $-1/7$. Continue. We are making a series with sum $0$, but all partial sums are negative. We can now partition the natural numbers into finite parts with sums $\ge 0$. The parts are now of different and growing sizes, but it's a very similar construction. $\endgroup$ Jul 24, 2012 at 4:24

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