# What's the limit of the series $\log_2(1-x)+x+x^2+x^4+x^8+\cdots$.

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

• Does the limit exist? Commented Jul 28, 2018 at 5:51
• Most probably yes....it's a decreasing function bounded from below. Commented Jul 28, 2018 at 7:27
• What does the last part, $$\sum_{k=0}^\infty x^{2^k}$$ alone give ($|x| < 1$)? Commented Jul 28, 2018 at 8:38
• How did you get that lower bound? Perhaps if you added that, the question might be reopened.
– robjohn
Commented Jul 31, 2018 at 18:03
• The original problem is: Prove that the function $f\colon\left[0,1\right)\to\mathbb{R}$ defined by $$f(x)=\log_2{\!(1-x)}+x+x^2+x^4+x^8+\dotsb$$ is bounded. See Problems from the Book (19.2, 11) by Titu Andreescu and Gabriel Dospinescu. Commented Sep 1, 2021 at 12:32

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.

• When you integrate (4), you average over the possible values of $f(1)$. My graph says $f(x)$ oscillates near $x=1$ with amplitude 0.0000015. Is that true, or a numerical artifact of underflow ? Commented Jul 29, 2018 at 7:32
• @Michael: if we differentiate $(3)$, and evaluate as $x\to1^-$, we get $f'(1)=\frac1{2\log(2)}-1$; so the function apparently decreases to $\frac12-\frac\gamma{\log(2)}$ as $x\to1^-$.
– robjohn
Commented Jul 29, 2018 at 12:09
• The periodicity is that $f(x)=p(x)+q(x)$ where $q(x)=q(x^2)$, for example $q(x)=\sin(2\pi\log_2(-\log(x)))$. So $q(x)$ is cancelled in (3), which therefore can't rule out $q(x)$ Commented Jul 29, 2018 at 12:24
• @Michael: ah, yes. I have encountered this before, and I was worried about it because of $(4)$. Equation $(3)$ kills the effect of any such $q$. I will have to think on this and try to see if I can compute this component.
– robjohn
Commented Jul 29, 2018 at 12:36
• It's not exact, but very close with $x=2^{-2^{-t}}$: $$\frac{1}{2} - \frac{\gamma}{\ln(2)} - \frac{\left(\pi^{\pi} + \gamma^{\pi}\right)^{{1}/{\pi}}}{2}\cdot 10^{-6} \, \sin(2\pi t - 0.692)$$ Commented Jul 30, 2018 at 22:59

$$f(x)+\log_2(1+x)-x=f(x^2)\text{ ( 0<x<1 )}\\ f(exp(y))+\log_2(1+\exp(y))-\exp(y)=f(\exp(2y))\text{ (-\infty<y<0)}\\ g(y)+\log_2(1+\exp(y))-\exp(y)=g(2y)\text{ (-\infty<y<0)}\\ g(-2^z)+\log_2(1+\exp(-2^z))-\exp(-2^z)=g(-2^{z+1})\text{( -\infty<z<\infty)}\\ h(z)+\log_2(1+\exp(-2^z))-\exp(-2^z)=h(z+1)\text{(-\infty<z<\infty)}\\ f(1)-f(0)=\int_{-\infty}^{\infty}dh=\int_{-\infty}^\infty\log_2(1+\exp(-2^z))-\exp(-2^z)dz\\ \approx -0.332746$$ By change of variable, it becomes $$\int_0^1 \frac{x-\log_2(1+x)}{x\log x\log 2}dx$$ which the Inverse Symbolic Calculator gives as $$\frac12-\frac\gamma{\ln2}\approx -0.3327461772769$$ As pointed out by Somos, I took an approximation when I replaced $\sum h(z+1)-h(z)$ by $\int dh$. It seems to have variation in the sixth decimal place as $x$ varies from $x_0$ to $x_0^2$.

Let $\ f(x) := \log_2(1-x) + \sum_{n=0}^\infty x^{2^n}, \$ $\ g(x) := f(e^{-x}) = \log_2(1-e^{-x}) + \sum_{n=0}^\infty e^{-x2^n}, \$ and $\ a_k := g(2^{-k}) = b_k + \sum_{n=0}^\infty e^{-2^{n-k}} \$ where $\ b_k := \log_2(1-e^{-2^{-k}}) \approx -k - 2^{-1-k}/\log(2). \$ Now $\ \sum_{n=0}^\infty e^{-2^{n-k}} = \sum_{n=1}^k e^{-2^{-n}} + B \$ where $\ B := \sum_{n=0}^\infty e^{-2^n} \approx 0.521865938459879089046726. \$ But $c_k :=\! -k \!+\! \sum_{n=1}^k e^{-2^{-n}}\! = \sum_{n=1}^k \big(e^{-2^{-n}}\!-\!1 \big) \$ and $\ c_k \to C \$ where $\ C \approx -0.8546133208927. \$ Finally, $\ \lim_{x\to 1^-} f(x) = \lim_{x\to 0^+} g(x) = \lim_{k\to\infty} a_k = B+C \approx -0.3327473824328992250.\$ The digits of $1$ minus this number is OEIS sequence A158468. $f(\exp(-2^{-30})) \approx -0.3327473822.$

EDIT: Unfortunately, it seems that the function $\ f(x) \$ oscillates as it gets close to $1$ from below. That is, $\ g(2^{-x}) \$ approaches a period $1$ function with mean value $\ 1/2 - \gamma/\log(2) \$ with oscillations of magnitude $\ \approx 1.57315\times 10^{-6} \$ as Michael shows. Thus, the limit does not exist. It was obvious that the infinite sum in $\ f(x) \$ has radius of convergence $1.$ What was not obvious was the limiting behavior as $\ x\to 1^-. \$ We now know that the series has a logarithmic singularity and $\ f(x) \$ is what remains. That $\ f(x) \$ has interesting oscillatory behavior is nice information.