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

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anonymous
  • anonymous
now I realize that I don't think I know what to do here :/
anonymous
  • anonymous
Is it equal 1/3?
Valpey
  • Valpey
Well, I'm pretty sure it is equal to 1. But I want to show it.
anonymous
  • anonymous
Yes, it is 1. I summed from n=1 to get 1/3
Valpey
  • Valpey
Still working on how to show it...
anonymous
  • anonymous
*
anonymous
  • anonymous
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.
Valpey
  • Valpey
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 \[a_n=\frac{1}{2^{(2^{n})}-2^{-(2^{n})}}=\frac{2^{(2^{n})}}{2^{(2^{n+1})}-1}\]
anonymous
  • anonymous
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
anonymous
  • anonymous
we are done....:)
anonymous
  • anonymous
Yes, we are.
TuringTest
  • TuringTest
@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
TuringTest
  • TuringTest
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} \[s(n-1)+ a_n=\frac{2^{2^n}-2 }{2^{2^n}-1}+\frac{2^{2^n}}{2^{2^{n +1}}-1}\]\[=\frac{2^{2^{n+1}}-2}{\left( 2^{2^n}-1\right) \left(2^{2^n}+1\right)}=\frac{2^{2^ {n+1}}-2}{2^{2^{n+1}}-1}\]
anonymous
  • anonymous
@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 \[ s(n-1)+ a_n=\frac{2^{2^n}-2 }{2^{2^n}-1}+\frac{2^{2^n}}{2^{2^{n +1}}-1}=\\\frac{2^{2^{n+1}}-2}{\left( 2^{2^n}-1\right) \left(2^{2^n}+1\right)}=\frac{2^{2^ {n+1}}-2}{2^{2^{n+1}}-1} \]
Valpey
  • Valpey
Very cool. This was fun. Thanks @eliassaab!
anonymous
  • anonymous
yw. @Valpey, You gave me the idea of the last step.
TuringTest
  • TuringTest
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.
anonymous
  • anonymous
The LaTeX preview appears on my browser perfect and never shift right or left on my browser. What browser are you using?
anonymous
  • anonymous
yeah its perfect on my screen...

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