Evaluate $\lim_{x\to1^-}\left(\sum_{n=0}^{\infty}\left(x^{(2^n)}\right)-\log_2\frac{1}{1-x}\right)$ 
Evaluate$$\lim_{x\to1^-}\left(\sum_{n=0}^{\infty}\left(x^{(2^n)}\right)-\log_2\frac{1}{1-x}\right)$$

Difficult problem. Been thinking about it for a few hours now. Pretty sure it's beyond my ability. Very frustrating to show that the limit even exists.
Help, please. Either I'm not smart enough to solve this, or I haven't learned enough to solve this. And I want to know which!
 A: Write $x:=e^{-2^{\delta}}$.  Then the desired limit is $\lim_{\delta\to-\infty} F(\delta)+\log_2 (1-e^{-2^{\delta}})$, where
$$
F(\delta):=\sum_{n\ge 0} e^{-2^{\delta+n}}.$$
But if
$$
G(\delta):=\sum_{n\ge 0} e^{-2^{\delta+n}}+\sum_{n<0} (e^{-2^{\delta+n}}-1)
$$
then shifting the index of summation shows that $G(\delta+1)=G(\delta)-1$, so $G(\delta)+\delta$ has period $1$.  Calling this periodic function $H(\delta)$,
then,
\begin{eqnarray*}
F(\delta)+\log_2 (1-e^{-2^{\delta}})
&=& H(\delta) -\delta + \log_2 (1-e^{-2^{\delta}}) - \sum_{n<0} (e^{-2^{\delta+n}}-1)\\
&=&H(\delta)+O(2^\delta),\qquad\delta\to-\infty.
\end{eqnarray*}
Computing the periodic function $H$ numerically shows that it is not a constant.  Therefore, the function whose limit is being taken is oscillatory, so the limit does not exist.
A: This approach is an addendum to David Moews' answer. Applying the method in Hardy's book 'Divergent Series' (4.10.2), it is possible to avoid the numerical approach. Denote by 
$$
F(x)=\sum_{n=0}^{\infty} x^{2^n}, \ \ F(x^2)=F(x)-x.
$$
Consider the function 
$$
\Phi(x)=\sum_{n=1}^{\infty}\frac{(\log x)^n}{(2^n-1)n!}, \ \ \Phi(x^2)=x-1+\Phi(x).
$$
On the other hand, 
$$
\log_2(\log \frac1{x^2})=1+\log_2(\log \frac1x).
$$
Then $G(x)=F(x)+\Phi(x)+\log_2(\log \frac1x)$ satisfies
$$
G(x^2)=G(x).
$$
Using principal branches of the logarithms, $G(z)$ is analytic on $\{z: |z|<1, \ \ z\notin (-1,0]\}$. Put $z=re^{i\pi/4}$, and let $r\rightarrow 1-$. Then we have
$$
|F(re^{i\pi/4})|\rightarrow\infty, \ \ 
$$
$$
|\Phi(re^{i\pi/4})| \textrm{ is bounded, }
$$
$$
\log_2(\log(\frac1z)) \textrm{ is bounded. }
$$
This shows that $G(z)$ cannot be a constant. Thus, $G(x)$ for $0<x<1$ is also not a constant. Hence, $\lim_{x\rightarrow 1-} G(x)$ is oscillatory and it does not exist. 
To finish up, consider
$$ 
\lim_{x\rightarrow 1-} \left( \log_2(\log \frac 1x)+\log_2 \frac1{1-x}\right) = \lim_{x\rightarrow 1-} \left( \log_2 \frac{\log \frac 1x}{1-x}    \right) =0,
$$
and
$$
\lim_{x\rightarrow 1-} \Phi(x) = 0.  
$$
Thus, it follows that 
$$
\lim_{x\rightarrow 1-} \left(\sum_{n=0}^{\infty} x^{2^n}-\log_2 \frac1{1-x} \right)=\lim_{x\rightarrow 1-} \left(G(x)-\Phi(x)- \left( \log_2(\log \frac 1x)+\log_2 \frac1{1-x}\right)\right) 
$$
is oscillatory and it does not exist. 
A: This question was re-asked recently: What's the limit of the series $\log_2(1-x)+x+x^2+x^4+x^8+\cdots$.
There it was shown that for $x\rightarrow 1^{-}$
$$
\lim_{x\to1^-}\left(\sum_{n=0}^{\infty}\left(x^{(2^n)}\right) + \log_2(1-x)\right)
$$
is actually a periodic function in the variable $t$ (for $t\rightarrow \infty$) with period $1$ where $x=e^{-2^{-t}}$. This periodic function for $t \rightarrow \infty$ can be written as
$$
f(t)=\sum_{k=-\infty}^\infty\left\{e^{-2^{k-t}}-\log_2\left(1+e^{-2^{k-t}}\right)\right\} \, .
$$
As such it can be expanded according to Fourier i.e.
$$
f(t)=\sum_{n=-\infty}^{\infty} c_n \, e^{i2\pi  nt}
$$
where
\begin{align}
c_0 &= \frac{1}{2} - \frac{\gamma}{\log 2} \\
c_n &= \frac{\Gamma\left(\frac{2\pi i \,n}{\log 2}\right)}{\log 2} \qquad n \neq 0 \, .
\end{align}
In terms of sine and cosine
$$
f(t)=c_0 + \sum_{n=1}^\infty \left\{ a_n \, \cos(2\pi nt) + b_n \, \sin(2\pi nt) \right\}
$$
where
\begin{align}
a_n &= c_n + c_{-n} = \frac{2\,{\rm Re}\left\{\Gamma\left(\frac{2\pi i \,n}{\log 2}\right)\right\}}{\log 2} \\
b_n &= i\left(c_n - c_{-n}\right) = -\frac{2\,{\rm Im}\left\{\Gamma\left(\frac{2\pi i \,n}{\log 2}\right)\right\}}{\log 2} \, .
\end{align}
In the amplitude-phase representation
$$
f(t) = c_0 + \sum_{n=1}^\infty A_n \, \cos\left(2\pi nt - \varphi_n\right)
$$
the amplitude becomes an elementary function by the identity
$$
\left|\Gamma\left(iz\right)\right|^2 = \frac{\pi}{z\, \sinh\left(\pi z\right)}
$$
and
\begin{align}
A_n &= \sqrt{a_n^2 + b_n^2} = \frac{2 \, \left|\Gamma\left(\frac{2\pi i \,n}{\log 2}\right)\right|}{\log 2} = \sqrt{\frac{2}{n \, \sinh\left(\frac{2\pi^2 n}{\log 2}\right) \, \log 2}} \\
\tan \varphi_n &= \frac{b_n}{a_n} = -\frac{{\rm Im}\left\{\Gamma\left(\frac{2\pi i \,n}{\log 2}\right)\right\}}{{\rm Re}\left\{\Gamma\left(\frac{2\pi i \,n}{\log 2}\right)\right\}} \\ \Longrightarrow \qquad -\varphi_n &= \arg\left\{ \Gamma \left(\frac{2\pi i \,n}{\log 2}\right) \right\} = -\frac{\pi}{2} - \frac{\gamma \, 2\pi \, n}{\log 2} + \sum_{k=1}^\infty \left\{\frac{2\pi \, n}{k\log 2} - \arctan \left( \frac{2\pi \, n}{k\log 2} \right) \right\} \, .
\end{align}
$A_2/A_1 \approx 4.63 \cdot 10^{-7}$ such that the first harmonic is already an excellent approximation.

Calculation of $c_n$: By definition
\begin{align}
c_n &= \int_0^1 f(t) \, e^{-i2\pi nt} \, {\rm d}t \\
&= \sum_{k=-\infty}^\infty \int_0^1 \left\{e^{-2^{k-t}} - \log_2\left(1+e^{-2^{k-t}}\right) \right\} e^{-i2\pi nt} \, {\rm d}t \, .
\end{align}
Substituting $u=2^{k-t}$, $\log u = (k-t)\log 2$, $\frac{{\rm d}u}{u} = -{\rm d}t \log 2$ leads to
\begin{align}
&= \sum_{k=-\infty}^\infty \int_{2^{k-1}}^{2^k} \left\{e^{-u} - \log_2\left(1+e^{-u}\right) \right\} u^{\frac{2\pi i \, n}{\log 2}-1} \, \frac{{\rm d}u}{\log 2} \\
&= \frac{1}{\log 2} \int_0^\infty \left\{e^{-u} - \log_2\left(1+e^{-u}\right) \right\} u^{\frac{2\pi i \, n}{\log 2}-1} \, {\rm d}u \, .
\end{align}
For $n=0$ partial integration leads to an integral representation for which the result $c_0$ given above is manifest. For $n\neq 0$ partial integration only in the second term gives
\begin{align}
c_n &= \frac{1}{\log 2} \int_0^\infty \left\{e^{-u} - \frac{u/(2\pi i \, n)}{e^{u} +1} \right\} u^{\frac{2\pi i \, n}{\log 2}-1} \, {\rm d}u \\
&= \frac{1}{\log 2} \left\{ \Gamma\left(\frac{2\pi i \, n}{\log 2} \right) - \frac{\Gamma\left(1 + \frac{2\pi i \, n}{\log 2} \right) \eta\left(1 + \frac{2\pi i \, n}{\log 2} \right) }{2\pi i \, n} \right\}
\end{align}
where $\eta(s)$ is the Dirichlet $\eta$-function. Using $$\eta(s) = \left(1-2^{1-s}\right) \zeta(s)$$ it is readily seen, that the $\log_2$-term of the series does not contribute.
A: This is NOT a solution, but I think that others can benefit from my failed attempt. Recall that $\log_2 a=\frac{\log a}{\log 2}$, and that $\log(1-x)=-\sum_{n=1}^\infty\frac{x^n}n$ for $-1\leq x<1$, so your limit becomes
$$\lim_{x\to1^-}x+\sum_{n=1}^\infty\biggl[x^{2^n}-\frac1{\log2}\frac{x^n}n\biggr]\,.$$
The series above can be rewritten as $\frac1{\log2}\sum_{k=1}^\infty a_kx^k$, where
$$a_k=\begin{cases} -\frac1k,\ &\style{font-family:inherit;}{\text{if}}\ k\ \style{font-family:inherit;}{\text{is not a power of}}\ 2;\\\log2-\frac1k,\ &\style{font-family:inherit;}{\text{if}}\ k=2^m.\end{cases}$$
We can try to use Abel's theorem, so we consider $\sum_{k=1}^\infty a_k$. Luckily, if this series converges, say to $L$, then the desired limit is equal to $1+\frac L{\log2}\,$. Given $r\geq1$, then we have $2^m\leq r<2^{m+1}$, with $m\geq1$. Then the $r$-th partial sum of this series is equal to
$$\sum_{k=1}^ra_k=\biggl(\sum_{k=1}^r-\frac1k\biggr)+m\log2=m\log2-H_r\,,$$
where $H_r$ stands for the $r$-th harmonic number. It is well-known that
$$\lim_{r\to\infty}H_r-\log r=\gamma\quad\style{font-family:inherit;}{\text{(Euler-Mascheroni constant)}}\,,$$
so $$\sum_{k=1}^ra_k=\log(2^m)-\log r-(H_r-\log r)=\log\Bigl(\frac{2^m}r\Bigr)-(H_r-\log r\bigr)\,.$$
Now the bad news: the second term clearly tends to $-\gamma$ when $r\to\infty$, but unfortunately the first term oscillates between $\log 1=0$ (when $r=2^m$) and $\bigl(\log\frac12\bigr)^+$ (when $r=2^{m+1}-1$), so $\sum_{k=1}^\infty a_k$ diverges.
