Originally posted by David113to be exact,
If the numerators are the number of even digits is the denominator, the sum is an ugly transcendental number.
-1/9+ \sum_{n=1}^{\infty} \frac { floor(n log_{10} 2)+1) }{2^n}
We want LaTeX !!!
Those who agree, come to the "site ideas" forum and make some pression!
You can start by considering another way to write a number.
For example:
0.532 = 0.5 + 0.03 + 0.002
Any number can be written as \sum_{k=1}^{\infty} \frac{a_k}{10^k}
For the number 0.532, a_1 =5 ; a_2=3; a_3 = 2
So, a_k gives the k-th digit of a given number
But there's another way to write a_k:
a_k = floor ( 0.532*10^k ) mod(10)
now instead, of base 10, try base 2, and find a relation between both 🙂
Originally posted by JirakonWrite the decimal numbers 0.1, 0.01, 0.001, 0.0001, ... in binary:
So what's the proof?
0.1 = 0.xyz...
0.01 = 0.abc...
0.001 = 0.pqr...
...
Then sum the binary fractions column by column, i. e. their sum is (x+a+p+...)/2+(y+b+q+...)/4+(z+c+r+...)/8+...
Show that this last series is exactly our series.