Pseudo-alternate series - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-26T02:22:27Z http://mathoverflow.net/feeds/question/56040 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/56040/pseudo-alternate-series Pseudo-alternate series Beni Bogosel 2011-02-20T05:46:44Z 2011-02-21T05:16:58Z <p>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 $\lim\limits_{n\to \infty}(\varepsilon_1+\varepsilon_2+...+\varepsilon_n) a_n=0$?</p> http://mathoverflow.net/questions/56040/pseudo-alternate-series/56095#56095 Answer by Igor Rivin for Pseudo-alternate series Igor Rivin 2011-02-20T19:10:59Z 2011-02-20T19:10:59Z <p>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.</p> <p><strong>EDIT</strong> however, if in the OP you replace $\lim$ by $\liminf,$ I am not sure what the answer is.</p> http://mathoverflow.net/questions/56040/pseudo-alternate-series/56141#56141 Answer by fedja for Pseudo-alternate series fedja 2011-02-21T05:16:58Z 2011-02-21T05:16:58Z <p>That would be more than welcome on AoPS, College Playground. For MO, it is hardly appropriate. </p> <p>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 $0$. </p>