Positivity of "harmonic" summation The settings for the problem are as follows. Given
a real number $\alpha\in[0,1]$, consider
a sequence of real (positive, negative and zero) numbers
$a_1,a_2,\dots,a_n,\dots$ satisfying
(1) $a_1=1$,
(2) $|a_n|\le n^\alpha$ for all $n=1,2,\dots$, and
(3) $\displaystyle\max_{1\le k\le n}\lbrace a_1+a_2+\dots+a_k\rbrace
+\min_{1\le k\le n}\lbrace a_1+a_2+\dots+a_k\rbrace\ge0$ again for all $n=1,2,\dots$.
Is it true that
$$
s_n=\sum_{k=1}^n\frac{a_k}k>0
$$
for any $n$?
The answer is no for $\alpha=1$, as the choice $a_k=(-1)^{k-1}k$ shows
that we can only achieve a nonstrict inequality in this case. So, what
are the conditions on $\alpha$ to ensure $s_n>0$ for any $n$?
I spent some time trying to construct a counterexample (for $\alpha=0$
and $\alpha=1/2$) but with no result. Let me note that one can
consider a finite sequence $a_1,a_2,\dots,a_n$ (but of arbitrary
length $n$, of course) which corresponds to the choice
$a_{n+1}=a_{n+2}=\dots=0$. A tedious analysis shows that $\alpha<1$
implies $s_n>0$ for $n=1,2,3,4$ but sheds no light on how to proceed
further.
Any ideas?!
EDIT. The problem has finally got a
solution in negative in the most interesting case $\alpha=0$. (This is
automatically a solution for any $\alpha\ge 0$.)
 A: You can get counter examples for various $\alpha$ of the following form.  Let
$$a_1=1, a_2=-2 + 2 \epsilon, a_3=0, a_4=a, a_5 = -b.$$
Here $0<\epsilon <1$ and we take $a$ is big enough so that $a + 2\epsilon -1 \ge 1$. Here is one such counterexample
$$a_1 = 1, a_2 = -\frac{7}{4}, a_3=0, a_4=3, a_5= -\frac{71}{16}.$$
The sequence of partial sums is $ 1, -\frac{3}{4},-\frac{3}{4}, \frac{9}{4}, -\frac{35}{16}$ which satisfies your min-max requirement.  The harmonic sum is
$$1 -\frac{7}{8} + \frac{3}{4} - \frac{71}{80} = \frac{-1}{80}.$$
More generally, the constraints on $a,b$, $\epsilon$ are
$$ 2(a + 2\epsilon -1) - b >0$$
(from the max-min restriction), and 
$$ \epsilon + \frac{a}{4}- \frac{b}{5} <0 $$
(so we get a counter example). In other words,
$$ 5 \epsilon + \frac{5a}{4} < b <2a + 4 \epsilon -1.$$
For $a> \frac{8}{3} +  \frac{4}{3} \epsilon$ these are consistent and leave us room to choose $b$. (And $a$ automatically satisfies our previous requirement that $a>2+2\epsilon$.)
We also need $a <4$ and $b<5$ to fit your requirement that $|a_j| \le j^\alpha$.  The requirement $a<4$ only asks that $\epsilon <1$, which we already assumed, however $b<5$ forces $\epsilon <1/4$.  Thus only for $\epsilon <1/4$ can you find $a$ and $b$ which make the harmonic sum at order $5$ negative and satisfy all your constraints.
