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 $\alpha\ge 0$.)