1

I came across following bi-infinite sum:

$\sum_{k=-\infty}^{\infty} 2^k x^{2^k}$

Is this a known series? After some plotting I have the feeling that it could be equal or very similar to

$-\frac{1}{\ln(2) \ln(x)}$

for all $x\in(0,1)$. Is my assumption correct?

Did
  • 279,727
otmar
  • 344
  • I suggest the following edit to the title: either say "power of power of $x$ series" or just say "power of 2 series". – ahorn May 06 '16 at 20:03
  • 1
    I note that using the Cauchy Condensation Test on this series we get the known series $\sum_{k=-\infty}^{\infty} k x^{k},$ but this doesn't tell you what the actual sum of the original series is. Also, the Wikipedia page for Lacunary function and web searches for this term might turn up something. – Dave L. Renfro May 06 '16 at 20:20
  • Using Wolfram Alpha (see here and here we see that the values you mention are different, but indeed very close. Possibly $-1/\ln(2)\ln(x)$ describes the asymptotic behaviour as $x\rightarrow 0$. – Wojowu May 06 '16 at 20:31
  • How about: $\sum_{n=0}^\infty 2^nx^{2^n}$ converges iff $|x|<1$, while $\sum_{n=-\infty}^{-1} 2^nx^{2^n}$ converges for all $x$. – GEdgar May 06 '16 at 20:47
  • 1
    It is not equal to $-1/\log(2)\log(x)$ but could be close. – GEdgar May 06 '16 at 20:54

2 Answers2

1

Note that $f(x^2)=\frac12f(x)$, $x\in(0,1)$. Thus with $g(x)=1/f(e^{-x})$ we have $g(2x)=2g(x)$, $x\in(0,\infty)$. And with $h(x)=\ln g(e^x)-x$, $x\in\Bbb R$, we have $h(x+\ln 2)=h(x)$, which must therefore be periodic.

Numerically (summing $f$ only from $k=-50$ to $50$, but that should be pretty good when computing $h(x)$ for $0\le x\le\ln 2$), $h$ seems to make only a (sine-like?) oscillation between $\approx-0.3665228$ and $\approx-0.366503$ (both $\approx \ln\ln 2$). So if $h(x)=\ln\ln 2+\epsilon(x)$ with $\epsilon(x)\approx 0$, then $g(x)=e^{\ln\ln 2+\epsilon(\ln x)+\ln x}=v(x) x \ln 2$ with $\upsilon(x)=e^{\epsilon(\ln x)}\approx 1$, and then $$f(x)=\frac1{g(-\ln x)}=\frac{-1}{v(-\ln x)\ln 2\ln x}.$$ The numerical results seem to indicate that $|\epsilon(x)|<9.885\cdot 10^{-6}$ so that $|\upsilon(x)-1|<9.885\cdot 10^{-6}$, i.e., $$f(x)\approx -\frac1{\ln 2\ln x} $$ with a relative error $<10^{-5}$. On the other hand, the oscillation does not seem to disappear when computing more summands of $f$, which means that this bound for the relative error is pretty sharp.

Hagen von Eitzen
  • 374,180
-1

That sum can be written as $\sum_{k=-\infty}^\infty \left(2x^2\right)^k$ which I would then break into two parts: $\sum_{k=-\infty}^0 \left(2x^2\right)^k+ \sum_{k= 0}^\infty \left(2x^2\right)^k- 1$ (The -1 is there because k= 0 occurs twice and one has to be subtracted off.)

$\sum_{k= 0}^\infty \left(2x^2\right)^k$ is a geometric series with "common ratio" $2x^2$. That has sum $\frac{1}{1- 2x^2}$. $\sum_{k= -\infty}{0} \left(2x^2\right)^k= \sum_{k= 0}^\infty \left(\frac{1}{2x^2}\right)^k$ is also a geometric series with "common ratio" $\frac{1}{2x^2}$. That has sum $\frac{1}{1-\frac{1}{2x^2}}= \frac{2x^2}{2x^2- 1}$

user247327
  • 18,710