Prove it

This is F--ing difficult, but I'll get it done.
And for even numbers, too.

Originally posted by serigado
This is F--ing difficult, but I'll get it done.
And for even numbers, too.

If the numerators are the number of even digits is the denominator, the sum is an ugly transcendental number.

Originally posted by David113
If the numerators are the number of even digits is the denominator, the sum is an ugly transcendental number.

to be exact,
-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!

I got it.
Impossible to post without LaTeX 🙂

Originally posted by serigado
I got it.
Impossible to post without LaTeX 🙂

Can you sketch the basic idea? I couldn't think of a good way to count the number of odd digits in the powers of 2...

Originally posted by ilywrin
Can you sketch the basic idea? I couldn't think of a good way to count the number of odd digits in the powers of 2...

Can you do it for base 2 ? (count the 0 and 1 ? )

Originally posted by serigado
Can you do it for base 2 ? (count the 0 and 1 ? )

I'm not sure I follow you - 2^n in binary is written as 1 followed by exactly n zeros, so this is not very informative. Are you doing some transformation, or something?

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 🙂

*bump*

Originally posted by eldragonfly
*bump*

Every now and then this problem comes to my mind but I can't solve it. I want to solve it in a way that I don't use the result and prove its validity. I want to really calculate the sum. 😠

So what's the proof?

Originally posted by Jirakon
So what's the proof?

Write the decimal numbers 0.1, 0.01, 0.001, 0.0001, ... in binary:

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.