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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.

flag	asked May 11 2010 at 1:37 Henry Yuen 637●4●8

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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.) – Qiaochu Yuan May 11 2010 at 1:44

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? – Henry Yuen May 11 2010 at 2:41

The partial sums grow really, really, really slowly. What exactly do you want to know? – Qiaochu Yuan May 11 2010 at 2:59

Well, in particular I'm looking for an upper bound that's tighter than 1/(1-a). – Henry Yuen May 11 2010 at 3:25

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. – Qiaochu Yuan May 11 2010 at 3:37

Wadim Zudilin · Accepted Answer · 2010-05-11 02:55:19Z UTC

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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 $f(z)$.

answered May 11 2010 at 2:55

Wadim Zudilin
8,674●1●24●59

	I think a simpler motivation is that the function has a natural boundary. Anything one might naively call an elementary function won't. – Qiaochu Yuan May 11 2010 at 3:01
	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 $f(z)$). – Wadim Zudilin May 11 2010 at 3:10
	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. – Kevin O'Bryant May 11 2010 at 21:37
	Okay, now I see that the question includes $0<a<1$. – Kevin O'Bryant May 11 2010 at 21:37

Sum of subset of geometric series: a^2^n

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