Is it true that $\lceil (3/2)^n \rceil - (3/2)^n > (3/4)^n$ for all $n>1$? It is easy to check that
$$
\lceil (3/2)^n \rceil - (3/2)^n \ge 1/2^n
$$
(to see this, just look at the binary representation of $(3/2)^n$).
Prove or disprove the stronger statement that
$$
\lceil (3/2)^n \rceil - (3/2)^n > (3/4)^n
$$
for all integer $n>1$. Here $\lceil x \rceil$ is the ceiling function; e.g.
$\lceil 3/2 \rceil = 2$.
Note: Using PARI/GP, the statement has been verified up to $n=20000$. (For this verification, default precision may need to be increased and/or the original inequality should be restated in terms of the fractional part $\{(3/2)^n\}$, as Gottfried Helms explains below.)

From $3^n=2^n\cdot k+ r, 0\leq r<2^n-1$
$$\left \lceil \frac{3^n}{2^n} \right \rceil - \frac{3^n}{2^n} > \frac{3^n}{2^{2n}} \iff 
2^n\left \lceil \frac{3^n}{2^n} \right \rceil - 3^n > \frac{3^n}{2^n} \iff \\
2^n\left(\left \lfloor \frac{3^n}{2^n}\right \rfloor+1\right)-3^n>\frac{3^n}{2^n} \iff 
2^n(k+1)-3^n> \frac{3^n}{2^n}  \iff ...$$
but $3^n+2^n=2^n\cdot(k+1)+r$, thus
$$...\iff 2^n-r > \frac{3^n}{2^n} \tag{1}$$
It is worth noting that $r=3^n-2^n\left \lfloor \frac{3^n}{2^n}\right \rfloor$ and it is known that
$$r>2^n-\left \lfloor \frac{3^n}{2^n}\right \rfloor-2$$
admits only finitely many solutions for $n$ (although, none known). I.e. from some large enough $n_0$ onwards 
$$r\leq 2^n-\left \lfloor \frac{3^n}{2^n}\right \rfloor-2 \Rightarrow
2^n-r\geq 2^n - 2^n+\left \lfloor \frac{3^n}{2^n}\right \rfloor+2=\\
\left \lfloor \frac{3^n}{2^n}\right \rfloor+2 > \frac{3^n}{2^n}$$
and, via the equivalences from $(1)$, we have that the original inequality is valid, from some $n_0$ onwards.
