## A plausible positivity

After getting stuck with the previous positivity (it probably sounds too complex), I would like to give a version of the problem which is of most interest to me.

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 infinitely many terms of the sequence are nonzero. These settings and Dirichlet's convergence test guarantee that the series $$\sum_{n=1}^\infty\frac{a_n}n$$ converges.

Assume, in addition, that $$\max_{1\le k\le n}A_k+\min_{1\le k\le n}A_k\ge0 \qquad\text{for all}\quad n=1,2,\dots.$$

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.)

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 All right, but otherwise $|a_n|\le a=0$, all terms are zero.. – Wadim Zudilin Jun 19 2010 at 12:38 I haven't seen your first edit, sorry. – Andrey Rekalo Jun 19 2010 at 12:41 Is it known that the sum is always nonnegative? – Andrey Rekalo Jun 19 2010 at 13:11 No, even this isn't known. – Wadim Zudilin Jun 19 2010 at 13:15

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.

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:

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 $2/13-1/10>1/20$.

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 $1/100$.

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!).

 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.

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But how do we get $|A_n|\leq A$ ? – Pietro Majer Jun 19 2010 at 18:49
Define $A=\max|A_n|$. Recall that it is a finite sequence. – Sergei Ivanov Jun 19 2010 at 19:01
Indeed, we should have seen from the start that the boundedness of $|A_n|$ is a red herring because you could always achieve that by cutting off the tail of the sequence. (I didn't spot it either.) – Harald Hanche-Olsen Jun 19 2010 at 20:09
(But it is of course needed for convergence.) – Harald Hanche-Olsen Jun 19 2010 at 20:10
The role of $A$ is that a lower bound in terms of it was asked. It is indeed positive if $A$ is not too large (as can be seen e.g. from Abel transform). – Sergei Ivanov Jun 19 2010 at 20:26
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