# 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!

• 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 $\lim_{n \to \infty} n - \log_2 n$ – Santosh Linkha Feb 6 '13 at 6:29

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.

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.

• Perhaps $\sum a_k$ is Cesàro summable? Abel's theorem would still apply then. – Antonio Vargas Feb 6 '13 at 7:18
• @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. – Matemáticos Chibchas Feb 10 '13 at 1:04
• 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. – Antonio Vargas Feb 10 '13 at 2:02