# Thread: hard hard

1. ## hard hard

thank you CaptainBlack
help me how latex and prove
$\displaystyle \sqrt{1+2\sqrt{1+3\sqrt{1+4\sqrt{1+...}}}} = 3$

2. Originally Posted by perash
thank you CaptainBlack
help me how latex and prove
$\displaystyle \sqrt{1+2\sqrt{1+3\sqrt{1+4\sqrt{1+...}}}} = 3$
we use [tex] [ /math] here

3. Originally Posted by perash
thank you CaptainBlack
help me how latex and prove
$\displaystyle \sqrt{1+2\sqrt{1+3\sqrt{1+4\sqrt{1+...}}}} = 3$
$\displaystyle \sqrt{1+2\sqrt{1+3\sqrt{1+4\sqrt{1+...}}}} = 3$

4. It never becomes 3.
It only tends to become 3.

5. Originally Posted by perash
thank you CaptainBlack
help me how latex and prove
$\displaystyle \sqrt{1+2\sqrt{1+3\sqrt{1+4\sqrt{1+...}}}} = 3$
Try this link

it should give you some ideas. the proof isn't full, but maybe you can pick up the slack using induction or something to prove any claim you make

6. Originally Posted by Jhevon
Try this link

it should give you some ideas. the proof isn't full, but maybe you can pick up the slack using induction or something to prove any claim you make
That's a fascinating expansion.

-Dan

7. Originally Posted by topsquark
That's a fascinating expansion.

-Dan
it is, isn't it?! It's so simple to understand, but i never would have thought about doing that in a million years

8. Originally Posted by topsquark
That's a fascinating expansion.
Originally Posted by Jhevon
it is, isn't it?! It's so simple to understand, but i never would have thought about doing that in a million years
The poster on the other site who solved this problem claims it belongs to Ramanajuan.

If you never read anything of Ramanajuan, be prepared! Ramanajuan does this type of stuff of the all time. He realizes the most crazy manipulations.