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

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.

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.

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.