# 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 $\alpha\ge 0$.)

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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.

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I forgot to add the (probably obvious) statement that the $\alpha$ for which you get a counterexample depends on $\epsilon$ in a way which one needs to compute. –  Jeff Schenker Jun 16 '10 at 16:37
Thank you, Jeff. I'll try to follow your example to control the size of possible $\alpha$. For the moment, it's not clear to me whether your example works for $\alpha\le 1/2$ but probably your "pattern" can be continued further to decrease $\alpha$. –  Wadim Zudilin Jun 16 '10 at 21:22
Jeff, the best I can get with your example is $\alpha>0.855\dots$. There is no serious win when I slightly generalize by taking $a_3=\dots=a_{n-2}=0$, $a_{n-1}=a$ and $a_n=-b$. But moving $a_2=-2+2\epsilon$ farer (so that $a_2=\dots=a_{k-1}=0$ while $a_k=-2+2\epsilon$) will probably give any $\alpha>0$. So, what can be said for $\alpha=0$? (This is the most interesting case to me.) –  Wadim Zudilin Jun 16 '10 at 23:20
Jeff, even this generalized construction does not allow me to get something better than $\alpha>0.85$. Further ideas are needed... –  Wadim Zudilin Jun 17 '10 at 0:48