## Valpey Group Title $\sum_{n=0}^{\infty}\frac{1}{2^{(2^n)}-2^{-(2^n)}}$ one year ago one year ago

1. Valpey Group Title

$\Sigma_{n=0}^{\infty}{\frac{1}{2^{(2^n)}-2^{-(2^n)}}}$ $\Sigma_{n-0}^\infty{\frac{2^{(2^n)}}{2^{(2^n)}*2^{(2^n)}-1}}=\Sigma_{n-0}^\infty{\frac{2^{(2^n)}}{(2^{(2^n)}+1)(2^{(2^n)}-1)}}=\Sigma_{n-0}^\infty{\frac{1/2}{2^{(2^n)}+1}+\frac{1/2}{2^{(2^n)}-1}}$

2. colorful Group Title

you should write \sum instead of \Sigma in latex to make it prettier

3. colorful Group Title

$\sum_{n=0}^\infty{\frac{2^{(2^n)}}{2^{(2^n)}*2^{(2^n)}-1}}=\sum_{n=0}^\infty{\frac{2^{(2^n)}}{(2^{(2^n)}+1)(2^{(2^n)}-1)}}=\sum_{n=0}^\infty{\frac{1/2}{2^{(2^n)}+1}+\frac{1/2}{2^{(2^n)}-1}}$

4. colorful Group Title

now I realize that I don't think I know what to do here :/

5. eliassaab Group Title

Is it equal 1/3?

6. Valpey Group Title

Well, I'm pretty sure it is equal to 1. But I want to show it.

7. eliassaab Group Title

Yes, it is 1. I summed from n=1 to get 1/3

8. Valpey Group Title

Still working on how to show it...

9. mukushla Group Title

*

10. eliassaab Group Title

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.

11. Valpey Group Title

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 $a_n=\frac{1}{2^{(2^{n})}-2^{-(2^{n})}}=\frac{2^{(2^{n})}}{2^{(2^{n+1})}-1}$

12. eliassaab Group Title

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

13. mukushla Group Title

we are done....:)

14. eliassaab Group Title

Yes, we are.

15. TuringTest Group Title

@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

16. TuringTest Group Title

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}$

17. eliassaab Group Title

@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 $s(n-1)+ a_n=\frac{2^{2^n}-2 }{2^{2^n}-1}+\frac{2^{2^n}}{2^{2^{n +1}}-1}=\\\frac{2^{2^{n+1}}-2}{\left( 2^{2^n}-1\right) \left(2^{2^n}+1\right)}=\frac{2^{2^ {n+1}}-2}{2^{2^{n+1}}-1}$

18. Valpey Group Title

Very cool. This was fun. Thanks @eliassaab!

19. eliassaab Group Title

yw. @Valpey, You gave me the idea of the last step.

20. TuringTest Group Title

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.

21. eliassaab Group Title

The LaTeX preview appears on my browser perfect and never shift right or left on my browser. What browser are you using?

22. mukushla Group Title

yeah its perfect on my screen...