TITLE: What's the limit of the series $\log_2(1-x)+x+x^2+x^4+x^8+\cdots$.
QUESTION [13 upvotes]: Find
  $$\lim_{x\to1^-}\log_2(1-x)+x+x^2+x^4+x^8+\cdots$$

I have found $1-\dfrac{1}{\ln2}$ as a lower bound, but not further than that

REPLY [8 votes]: Let
$$
f(x)=\log_2(1-x)+\sum_{k=0}^\infty x^{2^k}\tag1
$$
then $f(0)=0$ and
$$
f\!\left(x^2\right)=\log_2\left(1-x^2\right)+\sum_{k=1}^\infty x^{2^k}\tag2
$$
and therefore,
$$
f(x)-f\!\left(x^2\right)=x-\log_2(1+x)\tag3
$$
Thus, for $x\in(0,1)$,
$$
\begin{align}
f(1)
&=f(1)-f(0)\\[12pt]
&=\sum_{k=-\infty}^\infty\left[f\!\left(x^{2^k}\right)-f\!\left(x^{2^{k+1}}\right)\right]\\
&=\sum_{k=-\infty}^\infty\left(x^{2^k}-\log_2\left(1+x^{2^k}\right)\right)\tag4
\end{align}
$$
Expanding $\log(1+x)$ into its Taylor Series in $x$, we get
$$
\begin{align}
\int_0^1x^{a-1}\log(1+x)\,\mathrm{d}x
&=\int_0^1\sum_{k=1}^\infty\frac{(-1)^{k-1}x^{a-1+k}}k\,\mathrm{d}x\\
&=\sum_{k=1}^\infty\frac{(-1)^{k-1}}{k(k+a)}\\
&=\frac1a\sum_{k=1}^\infty(-1)^{k-1}\left(\frac1k-\frac1{k+a}\right)\\
&=\frac1a\left(\sum_{k=1}^\infty\left(\frac1k-\frac1{k+a}\right)-2\sum_{k=1}^\infty\left(\frac1{2k}-\frac1{2k+a}\right)\right)\\
&=\frac1a\left(\sum_{k=1}^\infty\left(\frac1k-\frac1{k+a}\right)-\sum_{k=1}^\infty\left(\frac1k-\frac1{k+a/2}\right)\right)\\[3pt]
&=\frac{H(a)-H(a/2)}a\tag5
\end{align}
$$
where $H(a)$ are the Extended Harmonic Numbers. Apply $(5)$ to get
$$
\int_0^1\log_2\left(1+x^{2^k}\right)\,\mathrm{d}x=\frac{H\!\left(2^{-k}\right)-H\!\left(2^{-k-1}\right)}{\log(2)}\tag6
$$
Integration of a monomial gives
$$
\int_0^1x^{2^k}\,\mathrm{d}x=\frac1{2^k+1}\tag7
$$
Integrating $(4)$ over $[0,1]$ and using $(6)$ and $(7)$ yields
$$
\begin{align}
f(1)
&=\lim_{n\to\infty}\left[\sum_{k=-n}^n\frac1{2^k+1}-\frac{H\!\left(2^n\right)-H\!\left(2^{-n-1}\right)}{\log(2)}\right]\\
&=\lim_{n\to\infty}\left[\frac12+n-\frac{\gamma+n\log(2)+O\!\left(2^{-n}\right)}{\log(2)}\right]\\[3pt]
&=\frac12-\frac{\gamma}{\log(2)}\tag8
\end{align}
$$

Problem with the Use of Equation $\boldsymbol{(3)}$
As pointed out by Michael, the use of equation $(3)$ above ignores the fact that
$$
g(x)-g\!\left(x^2\right)=0\tag9
$$
does not mean $g(x)=0$. In fact, for any $1$-periodic $h$, i.e. $h(x)=h(x+1)$,
$$
g(x)=h\!\left(\log_2(-\log(x))\right)\tag{10}
$$
satisfies $(9)$. I have encountered this misbehavior before in Does the family of series have a limit? and Find $f'(0)$ if $f(x)+f(2x)=x\space\space\forall x$.
Thus, the value given in $(8)$ is an average of the values of $f(1)$ given by $(4)$.
The function given in $(4)$ for $x=2^{-2^{-t}}$ has period $1$ in $t$. I have computed $f(1)$ from $(4)$ for $x\in\left[\frac14,\frac12\right]$; that is, the full period $t\in[-1,0]$. I get a plot very similar to that of Michael:

which oscillates between $-0.33274775$ and $-0.33274460$. The horizontal line is
$$
\frac12-\frac\gamma{\log(2)}=-0.33274618
$$
which is pretty close to the average of the minimum and maximum.
I am still looking for an a priori method to compute this oscillation.