Question

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

Answer 1

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

Answer 2

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

Answer 3

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

Answer 4

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

Answer 5

Is it equal 1/3?

Answer 6

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

Answer 7

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

Answer 8

Still working on how to show it...

Answer 9

*

Answer 10

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.

Answer 11

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

Answer 12

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

Answer 13

we are done....:)

Answer 14

Yes, we are.

Answer 15

@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

Answer 16

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}

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

Answer 17

@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}

\]

Answer 18

Very cool. This was fun. Thanks @eliassaab!

Answer 19

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

Answer 20

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.

Answer 21

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

Answer 22

yeah its perfect on my screen...