Prove that this function is bounded This is an exercise from Problems from the Book by Andreescu and Dospinescu.  When it was posted on AoPS a year ago I spent several hours trying to solve it, but to no avail, so I am hoping someone here can enlighten me.
Problem:  Prove that the function $f : [0, 1) \to \mathbb{R}$ defined by
$\displaystyle f(x) = \log_2 (1 - x) + x + x^2 + x^4 + x^8 + ...$
is bounded.
A preliminary observation is that $f$ satisfies $f(x^2) = f(x) + \log_2 (1 + x) - x$.  I played around with using this functional equation for awhile, but couldn't quite make it work.  

If anyone's interested, here's another approach.
First, observe that $$ \log{2} = \int_{2^r}^{2^{r+1}} \frac{1}{x}\;dx \le \sum_{k=2^r}^{2^{r+1}-1}\frac{1}{k} \le \int_{2^r-1}^{2^{r+1}-1} \frac{1}{x}\;dx = \log{2} + \log\left(1+\frac{1}{2^{r+1}-2}\right) $$ for $r\ge1$, so using $\log(1+x) \le x$ (for $x>-1$) and $2^{r+1} - 2 \ge 2^r$, we get $$ \log{2} \le \sum_{k=2^r}^{2^{r+1}-1} \frac{1}{k} \le \log{2} + \frac{1}{2^r} $$ for all $r\ge0$ (check directly for $r=0$).
Thus for $x\in[0,1)$, we have
$$\begin{align*}
\lvert f(x)\log{2}\rvert = \lvert\sum_{r\ge0} x^{2^r}\log{2} - \sum_{n\ge1} \frac{x^n}{n}\rvert
&= \lvert\sum_{r\ge0} x^{2^r}(\log{2} - \sum_{k=2^r}^{2^{r+1}-1}\frac{1}{k})  + \sum_{r\ge0}\sum_{k=2^r}^{2^{r+1}-1}\frac{x^{2^r} - x^k}{k} \rvert \\
&\le \sum_{r\ge0}\frac{x^{2^r}}{2^r} + \sum_{r\ge0}\sum_{k=2^r}^{2^{r+1}-1}\frac{x^{2^r} - x^k}{k} \\
&< 2 + \sum_{r\ge0}x^{2^r}(1-x)\sum_{k=2^r}^{2^{r+1}-1}\frac{1+x+\cdots+x^{k-2^r-1}}{k} \\
&\le 2 + (1-x)\sum_{r\ge0}x^{2^r}\sum_{k=2^r}^{2^{r+1}-1}\frac{k-2^r}{k} \\
&\le 2 + (1-x)\sum_{r\ge0}x^{2^r}(2^r\cdot 1) \\
&\le 2 + (1-x)(x + 2\sum_{r\ge1}\sum_{k=2^{r-1}+1}^{2^r}x^k) \\
&= 2 + x(1-x) + 2(1-x)\sum_{k\ge2} x^k \\
&= 2 + x + x^2,
\end{align*}$$ so we're done.
