MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Suppose $(a_n)$ is a non-increasing sequence of positive real numbers and $\varepsilon_i = \{\pm 1\},\ \forall i \in \mathbb{N}$ such that $\sum\limits_{i=1}^\infty \varepsilon_i a_i$ is convergent. Is it true that $\lim\limits_{n\to \infty}(\varepsilon_1+\varepsilon_2+...+\varepsilon_n) a_n=0$?

share|cite|improve this question
What's the context for this problem? For which sequences do you know this works? – Yemon Choi Feb 20 '11 at 6:05
This is clearly true for absolute convergent series. – Beni Bogosel Feb 20 '11 at 6:31
This looks like a homework problem. What have you tried to verify/disprove the contention? Gerhard "Ask Me About System Design" Paseman, 2011.02.19 – Gerhard Paseman Feb 20 '11 at 7:17
I agree it looks like a homework problem -- and the question doesn't even come with a list of things the poster managed to prove... – Julien Puydt Feb 20 '11 at 8:12
It is not a homework problem. I had the problem discussed with some of my friends during a math contest. We weren't able to prove or disprove it. This was two years ago... I haven't posted any details, because I didn't manage to get very far with it. It is obvious that $a_n \to 0$. Then if the sequence $(\varepsilon_1+..+\varepsilon_n)$ is bounded, the result is obvious. That's what I could get so far. I remember proving that if the sequence $(\varepsilon_1+..+\varepsilon_n)$ is bounded from below or above, the result also follows, but I don't remember exactly how I did that. – Beni Bogosel Feb 20 '11 at 10:47
up vote 7 down vote accepted

That would be more than welcome on AoPS, College Playground. For MO, it is hardly appropriate.

The statement is always true. Start with the fact that if the sums $\sum_{i=k}^m b_i$ ($1\le k\le m\le N$) are bounded by $\delta$ and $u_i$ is an increasing sequence of numbers on $[1,2]$, then the sums $\sum_{i=k}^m b_i u_i$ are bounded by $2\delta$. Now, split the set of indices into intervals of length $N_j$ such that the last $a$ in each interval is at most twice less than the first $a$ in each interval and the next $a$ is smaller. Let $A_j$ be the starting $a$ of the $j-th$ interval. The observation we made shows that the supremum of the sums of $\epsilon$'s over all subintervals of the $j$-th interval times $A_j$ is at most $2\delta_j$ where $\delta_j\to 0$ (tails get small). This tells us that we need only show that the limit is $0$ over the indices corresponding to the block beginnings. Now, what happens for that subsequence is that whatever product we had for $j$ gets divided by at least $2$ when we pass to $j+1$, after which we add at most $2\delta_j$. It remains to note that if you start with any number and do a sequence of steps each of which is division by 2 followed by adding a number that gets closer and closer to $0$, you will get closer and closer to $0$.

share|cite|improve this answer
Thank you. I will check the details later, but I suppose you are correct. :) – Beni Bogosel Feb 21 '11 at 12:01
I wrote a detailed version of your answer here:… – Beni Bogosel Mar 11 '12 at 13:48

No. If $\epsilon_i = 1,$ for all $i$ and $a_i = 0$ for $i$ not a perfect square, and $a_{k^2} = 1/k^2,$ then your limit is not zero, and the series is convergent. If you don't like $0,$ make the non-square $a_i$ some arbitrary small positive sequence.

EDIT however, if in the OP you replace $\lim$ by $\liminf,$ I am not sure what the answer is.

share|cite|improve this answer
Actually the original post asks for $a_i$ to be monotonic, which is similar in spirit to your liminf, I think. – JBL Feb 20 '11 at 19:43
@JBL: oops, didn't read it carefully... – Igor Rivin Feb 20 '11 at 20:31

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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