# Thread: using induction to prove...

1. ## using induction to prove...

Hey everyone i was doing some AP work on my own and couldnt figure out how to do this question with induction. i know the steps just cant figure out how to use it in this case. Any help would be appreciated

$
a(n) = \sqrt(2 +\sqrt 2+...+ \sqrt2)
$

"n roots"

not sure if i typed it properly, but all of the square roots are under the first one so its the square root of (2 + squareroot of 2 + squareroot of 2) "n" times...

then i have to:

use induction to prove the sequence is increasing
and prove if its bounded above by 2

2. Originally Posted by IrrationalPI3
Hey everyone i was doing some AP work on my own and couldnt figure out how to do this question with induction. i know the steps just cant figure out how to use it in this case. Any help would be appreciated

$
a_n = \sqrt{2 +\sqrt {2+ \sqrt{2+\cdots + \sqrt2}}}
$

"n roots"

not sure if i typed it properly, but all of the square roots are under the first one so its the square root of (2 + squareroot of 2 + squareroot of 2) "n" times...

then i have to:

use induction to prove the sequence is increasing
and prove if its bounded above by 2

your sequence can be written as $a_1=\sqrt{2}$ and $a_{n+1}=\sqrt{2+a_n}, \ n \geq 1.$ now an easy induction over $n$ shows that $0 then it follows that the sequence $\{a_n \}$ is increasing because

$a_{n+1}=\sqrt{2 + a_n} > a_n$ is equivalent to $(a_n + 1)(a_n - 2) < 0,$ which is obviously true because $0 < a_n < 2.$

3. $a_n=\sqrt{2+\sqrt{2+\ldots+\sqrt{2}}}$

We can observe that $a_n=\sqrt{2+a_{n-1}}, \ \forall n\geq 2$

$a_1=\sqrt{2}, \ a_2=\sqrt{2+\sqrt{2}}\Rightarrow a_1

Suppose that $a_{n-1}

$a_n-a_{n+1}=\sqrt{2+a_{n-1}}-\sqrt{2+a_n}=\frac{a_{n-1}-a_n}{\sqrt{2+a_{n-1}}+\sqrt{2+a_n}}<0$

Then $a_n

$a_1=\sqrt{2}<2$

Suppose that $a_n<2$.

Then $a_{n+1}=\sqrt{2+a_n}<\sqrt{2+2}=\sqrt{4}=2$