6
$\begingroup$

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 $\alpha\ge 0$.)

$\endgroup$
0

1 Answer 1

3
$\begingroup$

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.

$\endgroup$
4
  • $\begingroup$ 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. $\endgroup$ Jun 16, 2010 at 16:37
  • $\begingroup$ 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 $\alpha$. $\endgroup$ Jun 16, 2010 at 21:22
  • $\begingroup$ 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.) $\endgroup$ Jun 16, 2010 at 23:20
  • $\begingroup$ Jeff, even this generalized construction does not allow me to get something better than $\alpha>0.85$. Further ideas are needed... $\endgroup$ Jun 17, 2010 at 0:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.