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?
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}
$
 A: 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?
A: 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!
