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?


1 Answer 1


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 $\ge 0$.

  • $\begingroup$ 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
  • 1
    $\begingroup$ @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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.