A plausible positivity - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-19T11:51:59Z http://mathoverflow.net/feeds/question/28743 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/28743/a-plausible-positivity A plausible positivity Wadim Zudilin 2010-06-19T12:01:40Z 2010-06-19T20:13:26Z <p>After getting stuck with the <a href="http://mathoverflow.net/questions/28374" rel="nofollow">previous positivity</a> (it probably sounds too <a href="http://mathoverflow.net/questions/28612/do-names-given-to-math-concepts-have-a-role-in-common-mistakes-by-students/28614#28614" rel="nofollow">complex</a>), I would like to give a version of the problem which is of most interest to me.</p> <p>Consider a sequence of real numbers $a_1,a_2,\dots,a_n,\dots$ with absolute values bounded above by the first term $a_1=a>0$, which satisfies, for all $n=1,2,\dots$, $$|A_n|\le A \qquad\text{where}\quad A_n=a_1+a_2+\dots+a_n.$$ In addition, assume that <em>infinitely many terms of the sequence are nonzero</em>. These settings and <a href="http://en.wikipedia.org/wiki/Dirichlet%27s_test" rel="nofollow">Dirichlet's convergence test</a> guarantee that the series $$\sum_{n=1}^\infty\frac{a_n}n$$ converges.</p> <p>Assume, in addition, that $$\max_{1\le k\le n}A_k+\min_{1\le k\le n}A_k\ge0 \qquad\text{for all}\quad n=1,2,\dots.$$</p> <p>The problem is to show that $$\sum_{n=1}^\infty\frac{a_n}n>0$$ and to provide, in terms of $a$ and $A$, a lower (strictly positive) bound for the series. (The latter is optional, as I am not sure that such a bound exists.)</p> http://mathoverflow.net/questions/28743/a-plausible-positivity/28760#28760 Answer by Sergei Ivanov for A plausible positivity Sergei Ivanov 2010-06-19T16:49:19Z 2010-06-19T20:13:26Z <p>The sum $\sum a_n/n$ can be negative. Below I construct a finite sequence; one can always add a negligibly small tail to get infinitely many non-zeroes.</p> <p>Begin with $a_1=1$ and $a_2=-1$. This gives $A_2=0$ and the partial sum of the main series is $1-1/2=1/2$. Then, repeat 100 times the following procedure:</p> <p>Pick an integer $k$ larger than the length of the sequence so far. Extend $(a_n)$ by zeroes up to $n=10k-1$. Then set $a_n=1$ for all $n$ from $10k$ to $11k-1$ and $a_n=-1$ for all $n$ from $11k$ to $13k-1$. The $k$ ones contribute less than $1/10k$ each to the main series $\sum a_n/n$, and this is less than $1/10$ in total. The $2k$ negative ones contribute absolute value at least $1/13k$ each, this sums up to at least $2/13$ of negative amount. So the partial sum of the main series went down by at least $2/13-1/10>1/20$.</p> <p>But we have $A_n=-k$ now (for $n=13k-1$). To fix this, extend $(a_n)$ by a huge amount of zeroes, followed by $k$ ones, so that the contribution of these ones to the main sum is less than $1/100$. </p> <p>Now we extended the sequence so that the last $A_n$ is zero again but the partial sum of the main series went down by at least $1/30$. Choose the next $k$ and repeat (finitely many times!).</p> <p> Since the sequence is finite, one can define $A=\max|A_n|$ to satisfy the condition $|A_n|\le A$. The $\max+\min$ condition is immediate from the construction.</p>