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

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?

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 $g'(x)=2x =\dfrac{1}{\frac1{2\sqrt{x^2}}}$.

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

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

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?

• 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. Dec 28, 2017 at 1:02
• Yes. I agree. But who knows, maybe I missed something. Dec 28, 2017 at 1:19
• Good example, so I upvoted it. Perhaps some condition like $f(x)/x \to 0$ would do what I want. Dec 28, 2017 at 1:51
• Yes, maybe, although it is not necessary ($f(x)=x$ works). Dec 28, 2017 at 2:00
• 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. Dec 28, 2017 at 16:44

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!