After getting stuck with the
[previous positivity](https://mathoverflow.net/questions/28374)
(it probably sounds too [complex](https://mathoverflow.net/questions/28612/do-names-given-to-math-concepts-have-a-role-in-common-mistakes-by-students/28614#28614)),
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](http://en.wikipedia.org/wiki/Dirichlet%27s_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.)



A plausible positivity
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!).


[Edit] 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.


