The formula for 1 + a + a^2 + .... where 0 < a < 1 is $\frac{1}{1-a}$, but what if you wanted to sum only those where the exponent is a power of 2? That is,

$S = a + a^2 + a^4 + a^8 + \cdots$

I feel like this is an easy one but I just can't seem to find a closed expression for it, nor search for it on Google.

  • 7
    $\begingroup$ There really isn't one. This is an example of a lacunary function (en.wikipedia.org/wiki/Lacunary_function), and it's known more for its undesirable analytic properties than anything else. (Admittedly, it satisfies a nice functional equation.) $\endgroup$ Commented May 11, 2010 at 1:44
  • $\begingroup$ Interesting. Are there any characterizations about how slowly the partial sums grow with respect to the growth of the partial sums of the geometric series? $\endgroup$
    – Henry Yuen
    Commented May 11, 2010 at 2:41
  • $\begingroup$ The partial sums grow really, really, really slowly. What exactly do you want to know? $\endgroup$ Commented May 11, 2010 at 2:59
  • $\begingroup$ Well, in particular I'm looking for an upper bound that's tighter than 1/(1-a). $\endgroup$
    – Henry Yuen
    Commented May 11, 2010 at 3:25
  • $\begingroup$ You can truncate the series at any finite point and assume that it continues like a geometric series and that gives you a sequence of upper bounds a + a^2 + ... + a^{2^n}/(1 - a^{2^n}). For moderately large n and moderately small a the error in this approximation will be pretty small. $\endgroup$ Commented May 11, 2010 at 3:37

1 Answer 1


Mahler proved in the 1930s that the values of $f(z)=\sum_{n=0}^\infty z^{d^n}$, $d>1$ is an integer, are transcendental for any algebraic $z$ satisfying $0<|z|<1$. A related problem of transcendence of the function $f(z)$ was discussed in this question. This motivates nonexistence of simple formula like $1/(1-z)$ for $f(z)$.

  • 2
    $\begingroup$ I think a simpler motivation is that the function has a natural boundary. Anything one might naively call an elementary function won't. $\endgroup$ Commented May 11, 2010 at 3:01
  • $\begingroup$ Agreed. My point is that Mahler's series are already special functions (as far as I know $f(z)$ does not satisfy an algebraic differential equation but only the functional equation relating $f(z^d)$ to $f(z)$). $\endgroup$ Commented May 11, 2010 at 3:10
  • $\begingroup$ Well, it motivates the nonexistence of an algebraic formula valid over the complexes. There are other "closed expressions" that may qualify, particularly if Henry isn't thinking of this as a complex power series. For example, if the coefficients are in the binary field $F_2$, then the series is algebraic. $\endgroup$ Commented May 11, 2010 at 21:37
  • $\begingroup$ Okay, now I see that the question includes $0<a<1$. $\endgroup$ Commented May 11, 2010 at 21:37

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.