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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!

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  • $\begingroup$ just playing around, $\lim_{x \to 1^-}\left( \frac{\lim_{n \to \infty }\left( 1-x^{2n}\right) - (1-x^{2})\log_2 \left( 1 \over 1-x\right)}{1-x^{2}}\right)$, this seem to go like $\lim_{n \to \infty} n - \log_2 n$ $\endgroup$ – Santosh Linkha Feb 6 '13 at 6:29
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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.

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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.

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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.

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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.

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  • $\begingroup$ Perhaps $\sum a_k$ is Cesàro summable? Abel's theorem would still apply then. $\endgroup$ – Antonio Vargas Feb 6 '13 at 7:18
  • $\begingroup$ @AntonioVargas Thanks for the suggestion, though from David Moews' answer we can conclude that $\sum a_k$ is not Cesàro summable, either. By the way, where can I find a proof of this more general version of Abel's theorem? Almighty Wikipedia does not make any reference to this generalization. $\endgroup$ – Matemáticos Chibchas Feb 10 '13 at 1:04
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    $\begingroup$ I learned the result from Hardy's book Divergent Series. Unfortunately I don't have a copy handy so I can't point you to a particular page. $\endgroup$ – Antonio Vargas Feb 10 '13 at 2:02

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