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 $\lim\limits_{n\to \infty}(\varepsilon_1+\varepsilon_2+...+\varepsilon_n) a_n=0$?
Edit: Some years later I found the same problem in Problems and Theorems in Analysis I, Problem 139.
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 $0$.
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.
