Calculating the Sum of Odd Digits in Powers of 2: A Fascinating Number Property

[Zhivago]

$$\sum_{n=1}^{\infty} \frac{o(2^n)}{2^n} = \frac{1}{9}$$
where $$o(2^n)$$ is the number of odd digits of $$2^n$$.

Found it in
http://mathworld.wolfram.com/DigitCount.html
equation (9)

 

[uart]

That's a pretty amazing series. I found some more information about it in the following paper. See : http://www.jstor.org/view/00029890/di991774/99p0626c/0

 

[Tedjn]

Curses, JSTOR!

 

[Zhivago]

That's a pretty amazing series. I found some more information about it in the following paper. See : http://www.jstor.org/view/00029890/di991774/99p0626c/0

Fantastic! Thanks a lot!
Also found it in
Experimentation in Mathematics: Computational Paths to Discovery
By Jonathan M. Borwein, David H. Bailey, Roland
pag 14-15

here's a google link:
http://books.google.com/books?id=cs...over&sig=yE9mO3b-YA9lLjAq6Nt4ED4bn1g#PPA15,M1

I'm no mathematician, but found really interesting some of these strange number properties.

 

[uart]

...here's a google link:
http://books.google.com/books?id=cs...over&sig=yE9mO3b-YA9lLjAq6Nt4ED4bn1g#PPA15,M1

I'm no mathematician, but found really interesting some of these strange number properties.

Thanks for the link Zhivago. Yes that book provides a nice accessible proof of that summation. In the link I posted they only really hinted at how that series was handled but in your link they nail it (only really needing knowledge of geomeric series and modolu athrithmetic to follow it). Good stuff!