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This is an attempt to generalize this question, which is the case $f(n) = \sqrt{n}$:

Boundedness of partial sums

Suppose $f(x) > 0, f'(x) > 0, f''(x) < 0, f(x) \to \infty $.

When is $\sum_{n=1}^{\infty} (-1)^{\lfloor f(n) \rfloor } $ bounded?

My answer:

Let $g$ be the inverse function of $f$. Then the sum is bounded if and only if $g'(x)$ is bounded.

In the question that inspired this, $f(x) = \sqrt{x}$, so $g(x) = x^2$. Since $g'(x) = 2x$ is not bounded, the sum is not bounded.


Here is my analysis.

Let $g(m) =\min(k|f(k) \ge m) $. Then $\lfloor f(n) \rfloor$ is constant for $g(m) \le n \lt g(m+1)$ and all such $f(i) = m$.

$g$ is an inverse function of $f$, so $g'(x) \approx \dfrac1{f'(g(x))} $ and $f'(x) \approx \dfrac1{g'(f (x))} $. For example, if $f(x) = \sqrt{x}$, $g(x) \approx x^2$ so $f'(x)=\dfrac1{2\sqrt{x}}$ and $g'(x)=2x =\dfrac{1}{\frac1{2\sqrt{x^2}}}$.

$\begin{array}\\ \sum_{n=1}^{g(m+1)-1} (-1)^{\lfloor f(n) \rfloor } &=\sum_{i=1}^m \sum_{j=g(i)}^{g(i+1)-1} (-1)^{i }\\ &=\sum_{i=1}^m (g(i+1)-g(i))(-1)^i\\ \text{so}\\ \sum_{n=1}^{g(2m+1)-1} (-1)^{\lfloor f(n) \rfloor } &=\sum_{i=1}^{2m} (g(i+1)-g(i))(-1)^i\\ &=\sum_{i=1}^{m} ((g(2i+1)-g(2i)-(g(2i)-g(2i-1)))\\ &=\sum_{i=1}^{m} (g(2i+1)-2g(2i)+g(2i-1)))\\ &\approx \sum_{i=1}^{m} g''(2i)\\ &\approx \frac12 g'(2m)\\ \end{array} $

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  • $\begingroup$ Since two good answers have been supplied, I will add my analysis to the question. $\endgroup$ Dec 28, 2017 at 1:53

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I don't it's true.

What if $f(x) = 10x + 3(1-\frac{1}{x})$?, $\lfloor{f(x)}\rfloor$ is clearly even after some point. And $f'(x) = 10 + \frac{3}{x^2}>0$ and $f''(x) < 0$, clearly. $f(x) \to \infty$, too. And $$g(x) = \frac{x-3}{20} + \frac{1}{20} \sqrt{x^2-6x+129}$$$$g'(x) = \frac{1}{20} + \frac{-6+2x}{40\sqrt{129 - 6 x + x^2}}$$Which is clearly bounded. Am I missing something?

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  • $\begingroup$ Yeah I had my doubts about this direction too. Intuitively it seems like a much more radical statement to me, so I'm not surprised if your conterexample holds. $\endgroup$ Dec 28, 2017 at 1:02
  • $\begingroup$ Yes. I agree. But who knows, maybe I missed something. $\endgroup$
    – Higurashi
    Dec 28, 2017 at 1:19
  • $\begingroup$ Good example, so I upvoted it. Perhaps some condition like $f(x)/x \to 0$ would do what I want. $\endgroup$ Dec 28, 2017 at 1:51
  • $\begingroup$ Yes, maybe, although it is not necessary ($f(x)=x$ works). $\endgroup$
    – Higurashi
    Dec 28, 2017 at 2:00
  • $\begingroup$ Yes good point. I think what makes this hard is $x^a$ for non integer $a$ greater than $1$. They would, to me, in general, seem to randomly fluctuate between odd and even integers when rounded down, so I'm not sure if all of these would be unbounded. $\endgroup$ Dec 28, 2017 at 16:44
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I will cover just one side of the coin here. The converse is left open.

I'll make a slightly different, but I believe equivalent, statement to the "if" side of your iff statement: If, as $x \to \infty$, $f'(x) \to 0$, then the sum will not be bounded.

Proof:

Let $$S(x) = \sum_{n=1}^{x} (-1)^{\lfloor f(n) \rfloor }$$

Because the slope gets arbitrarily small, we can always find a run of $2M$ values starting at some $m+1$ such that:

$$\lfloor f(m+1) \rfloor = \lfloor f(m+2) \rfloor \ldots = \lfloor f(m+2M) \rfloor$$

So:

$$S(m + 2M) = S(m) \pm 2M$$

There are then two cases to consider. Either the sum has already exceeded or equalled the bound $M$ at the point $m$, or it has not, in which case it is within the bound $(-m, m)$, exclusive.

So:

$$|S(m + 2M)| = |S(m) \pm 2M| \geq ||S(m)| - 2M| \geq |M - 2M| = M$$

So the bound is exceeded.

Showing the contrary is another piece of work I think!

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