Sum of subset of geometric series: a^2^n - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T02:06:54Z http://mathoverflow.net/feeds/question/24190 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/24190/sum-of-subset-of-geometric-series-a2n Sum of subset of geometric series: a^2^n Henry Yuen 2010-05-11T01:37:21Z 2010-05-11T02:55:19Z <p>The formula for 1 + a + a^2 + .... where 0 &lt; a &lt; 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,</p> <p>$S = a + a^2 + a^4 + a^8 + \cdots$ </p> <p>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. </p> http://mathoverflow.net/questions/24190/sum-of-subset-of-geometric-series-a2n/24197#24197 Answer by Wadim Zudilin for Sum of subset of geometric series: a^2^n Wadim Zudilin 2010-05-11T02:55:19Z 2010-05-11T02:55:19Z <p>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 $0&lt;|z|&lt;1$. A related problem of transcendence of the function $f(z)$ was discussed in <a href="http://mathoverflow.net/questions/21290/whats-an-example-of-a-transcendental-power-series/21554#21554" rel="nofollow">this question</a>. This motivates nonexistence of simple formula like $1/(1-z)$ for $f(z)$.</p>