# Math Help - [SOLVED] Just a challenge question

1. ## [SOLVED] Just a challenge question

What is the value of √(2+√(2+√(2+...))) etc?

This is just a challenge question for which I am curious about.

2. Hello,

Is it 2 ?

Assuming that $f(n+1)=\sqrt{2+f(n)}$

If L is the limit, then $\lim_{n \to \infty} f(n+1)=\lim_{n \to \infty} f(n)=L$

$L=\sqrt{2+L}$

--> $L^2=2+L \Longleftrightarrow L^2-L-2=0$

The solutions to this are -1 and 2. But L has to be > 0.

3. It could be...I have no idea how to check it. Can you explain how you worked that out? Thanks.

4. I edited

This is, of course, assuming that the series converges.

5. Thanks for that...

I, however, being an average tenth grader, do not understand that thing with the n+1 and the limit/infinity stuff. (basically, the entire explanation).

I'm thinking that the n+1 is something got to do with a series of numbers of something...but otherwise, I don't understand it.

If it helps, the challenge question provides a hint on how to start it:
Let x be √(2+√(2+√(2+...))) and square both sides.

If you could possibly explain it starting with a simpler method (for dummies), then perhaps I will be able to understand.

Thanks.

6. Oh, sorry for that

The n, n+1 thing means "the nth term of the sequence, etc..."

Well, with the hint, I can explain better (I hope), and it's quite alike with my method but it will be more clear

$x=\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{\dots}}}}$

Square it :

$x^2=2+{\color{red}\sqrt{2+\sqrt{2+\sqrt{\dots}}}}$

Since the dots mean "infinity", the red part is also equal to x (this is, imho, the most difficult part of the process, so if you understand it, it's great )

Hence $x^2=2+x$

$x^2-x-2=0$

The discriminant $\Delta=1+8=9$ (I don't know if you've already studied that, if not, complete the square )

Thus the solutions are $x_1=\dots$, $x_2=\dots$

But, since x is the square root of something, it has to be a positive number. This means that you can take only one of the solutions

7. Oh, I understand it now.

(PS: How much time do you spend on this forum? Every time I ask a question, it's always you who replies so quickly )

8. Originally Posted by anne.
Oh, I understand it now.