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\[\sum_{n=0}^{\infty}\frac{1}{2^{(2^n)}-2^{-(2^n)}}\]

Mathematics
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\[\Sigma_{n=0}^{\infty}{\frac{1}{2^{(2^n)}-2^{-(2^n)}}}\] \[\Sigma_{n-0}^\infty{\frac{2^{(2^n)}}{2^{(2^n)}*2^{(2^n)}-1}}=\Sigma_{n-0}^\infty{\frac{2^{(2^n)}}{(2^{(2^n)}+1)(2^{(2^n)}-1)}}=\Sigma_{n-0}^\infty{\frac{1/2}{2^{(2^n)}+1}+\frac{1/2}{2^{(2^n)}-1}}\]
you should write \sum instead of \Sigma in latex to make it prettier
\[\sum_{n=0}^\infty{\frac{2^{(2^n)}}{2^{(2^n)}*2^{(2^n)}-1}}=\sum_{n=0}^\infty{\frac{2^{(2^n)}}{(2^{(2^n)}+1)(2^{(2^n)}-1)}}=\sum_{n=0}^\infty{\frac{1/2}{2^{(2^n)}+1}+\frac{1/2}{2^{(2^n)}-1}}\]

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now I realize that I don't think I know what to do here :/
Is it equal 1/3?
Well, I'm pretty sure it is equal to 1. But I want to show it.
Yes, it is 1. I summed from n=1 to get 1/3
Still working on how to show it...
*
Let \[ s_n \] be the partial sum. See the first six partial sums \[ \left\{\frac{2}{3},\frac{14}{15},\frac{254}{255 },\\\frac{65534}{65535},\frac{4294967294}{4294 967295},\frac{18446744073709551614}{18446744 073709551615},\\\frac{340282366920938463463374 607431768211454}{340282366920938463463374607 431768211455}\right\} \] The numerator is 1 less than the denominator and this true for any at least numerically.
So I need to show that \[\large \text{Given }a_n=\frac{1}{2^{(2^{n})}-2^{-(2^{n})}}\]\[\large s_n=\frac{2^{2^{n+1}}-2}{2^{2^{n+1}}-1}\]I think I can start with \[a_n=\frac{1}{2^{(2^{n})}-2^{-(2^{n})}}=\frac{2^{(2^{n})}}{2^{(2^{n+1})}-1}\]
We can prove this by induction. It is true for n=1, suppose that it is true for n-1 \[ s(n-1)+ a_n=\frac{2^{2^n}-2 }{2^{2^n}-1}+\frac{2^{2^n}}{2^{2^{n +1}}-1}=\\\frac{2^{2^{n+1}}-2}{\left( 2^{2^n}-1\right) \left(2^{2^n}+1\right)}=\frac{2^{2^ {n+1}}-2}{2^{2^{n+1}}-1} \] @mukushla @Valpey
we are done....:)
Yes, we are.
@eliassaab Just a LaTeX note: I don't know how what you are typing looks on your screen, but you often type so that everything is shifted to the right (off the screen at times) I think you are putting in an extra \ when you make \\ that moves everything to the right, unless you are making a matrix or array
s(n-1)+ a_n=\frac{2^{2^n}-2 }{2^{2^n}-1}+\frac{2^{2^n}}{2^{2^{n +1}}-1}=\\\frac{2^{2^{n+1}}-2}{\left( ^^this part is shifting your work, try it with just the 1 slash and make separate brackets if necessary 2^{2^n}-1\right) \left(2^{2^n}+1\right)}=\frac{2^{2^ {n+1}}-2}{2^{2^{n+1}}-1} s(n-1)+ a_n=\frac{2^{2^n}-2 }{2^{2^n}-1}+\frac{2^{2^n}}{2^{2^{n +1}}-1}=\frac{2^{2^{n+1}}-2}{\left( 2^{2^n}-1\right) \left(2^{2^n}+1\right)}=\frac{2^{2^ {n+1}}-2}{2^{2^{n+1}}-1} \[s(n-1)+ a_n=\frac{2^{2^n}-2 }{2^{2^n}-1}+\frac{2^{2^n}}{2^{2^{n +1}}-1}\]\[=\frac{2^{2^{n+1}}-2}{\left( 2^{2^n}-1\right) \left(2^{2^n}+1\right)}=\frac{2^{2^ {n+1}}-2}{2^{2^{n+1}}-1}\]
@TuringTest, Thanks, I already know that. I do not see why you could not see We can prove this by induction. It is true for n=1, suppose that it is true for n-1 \[ s(n-1)+ a_n=\frac{2^{2^n}-2 }{2^{2^n}-1}+\frac{2^{2^n}}{2^{2^{n +1}}-1}=\\\frac{2^{2^{n+1}}-2}{\left( 2^{2^n}-1\right) \left(2^{2^n}+1\right)}=\frac{2^{2^ {n+1}}-2}{2^{2^{n+1}}-1} \]
Very cool. This was fun. Thanks @eliassaab!
yw. @Valpey, You gave me the idea of the last step.
I never said you can't prove it by induction funny, as I was typing this your work shifted back to the left... I was only trying to help you keep it from looking strange in latex, I said nothing of your mathematics.
The LaTeX preview appears on my browser perfect and never shift right or left on my browser. What browser are you using?
yeah its perfect on my screen...

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