## Pseudo-alternate series

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 $\lim\limits_{n\to \infty}(\varepsilon_1+\varepsilon_2+...+\varepsilon_n) a_n=0$?

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What's the context for this problem? For which sequences do you know this works? – Yemon Choi Feb 20 2011 at 6:05
This is clearly true for absolute convergent series. – Beni Bogosel Feb 20 2011 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 2011 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 2011 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 2011 at 10:47

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 $0$.

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 Thank you. I will check the details later, but I suppose you are correct. :) – Beni Bogosel Feb 21 2011 at 12:01 I wrote a detailed version of your answer here: mathproblems123.wordpress.com/2009/10/01/… – Beni Bogosel Mar 11 2012 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.

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Actually the original post asks for $a_i$ to be monotonic, which is similar in spirit to your liminf, I think. – JBL Feb 20 2011 at 19:43
@JBL: oops, didn't read it carefully... – Igor Rivin Feb 20 2011 at 20:31