TITLE: Lower bound for the fractional part of $(4/3)^n$
QUESTION [7 upvotes]: My son, who is 16, is doing some independent research. A lower bound depending on $n$ for $\left\{ \left( \frac{4}{3} \right)^n \right\}=\left( \frac{4}{3} \right)^n-\left\lfloor \left(\frac{4}{3} \right)^n \right\rfloor$, the fractional part of $\left( \frac{4}{3} \right)^n$, might help him improve his results. Note that $\frac{1}{3^n}$ is an obvious bound. Is there a better bound known?

REPLY [9 votes]: It may be interesting to note that, subject to the ABC conjecture, you have the fantastically good estimate
  $$ \left\{ \left( \frac43 \right)^n\right\} \gg_\delta \delta^n,\quad \delta\in(0,1). $$
The proof goes as follows. 
Let 
  $$ 4^n = 3^nk+r,\quad 0<r<3^n. $$
The greatest common divisor of $k$ and $r$ divides $4^n$, and we write 
  $$ (k,r)=2^d,\ k=2^dk_0,\ \text{and}\ r=2^dr_0, $$
so that $2^{-d}4^n=3^nk_0+r_0$. Assuming the ABC, for any $\varepsilon>0$ we have then
\begin{align*}
 2^{-d}4^n &\ll_\varepsilon \left( {\rm rad}(2^{-d}4^n\cdot3^nk_0\cdot r_0) \right)^{1+\varepsilon} \\
  &\le\ \ \ (6k_0r_0)^{1+\varepsilon} \\
  &\ll_\varepsilon (kr)^{1+\varepsilon}\cdot 2^{-d}.
\end{align*}
In view of $k<(4/3)^n$, this implies
  $$ 4^n \ll_{\varepsilon} \left(\frac43\right)^{(1+\varepsilon)n} r^{1+\varepsilon} $$
and, as a result,
  $$ \left\{ \left( \frac43 \right)^n \right\} = \frac r{3^n} \gg_{\varepsilon} 4^{-\varepsilon n/(1+\varepsilon)}. $$
The assertion follows by choosing $\varepsilon$ to satisfy $4^{-\varepsilon/(1+\varepsilon)}=\delta$.