What is the name and value of the constant \sum_{n=1}^\infty{2^{-2^n}}?

[Pere Callahan]

Hi,

I was wondering if the constant
<br /> \sum_{n=1}^\infty{2^{-2^n}}<br />

has a certain name or some history or anything. It certainly appears not to have a closed form expression. It also certainly has some value because it's majorized by the simple geometric series. It's numerical value is 0.31642150902189314371 (given by http://www.research.att.com/~njas/sequences/A078585" , but is there anything else known about it?

Regards,

Pere

 

[disregardthat]

Hi,

I was wondering if the constant
<br /> \sum_{n=1}^\infty{2^{-2^n}}<br />

has a certain name or some history or anything. It certainly appears not to have a closed form expression. It also certainly has some value because it's majorized by the simple geometric series. It's numerical value is 0.31642150902189314371 (given by http://www.research.att.com/~njas/sequences/A078585" , but is there anything else known about it?

Regards,

Pere

It's transcendental, if I'm not mistaken.
|\sum_{n=1}^\infty{2^{-2^n}}- \sum_{n=1}^k{2^{-2^n}}| = |\sum_{n=k+1}^\infty{2^{-2^n}}| = |\sum_{n=1}^\infty{2^{-2^n2^k}}| \leq |\sum_{n=1}^\infty{2^{-2^n}}|^{2^k} &lt; \left(\frac{1}{2}\right)^{2^k}=\frac{1}{2^{2^{k+1}}}

The denominator of the rational number \sum_{n=1}^k{2^{-2^n}} is 2^{2^k}. The number is thus a liouville number, and therefore transcendental.