# using induction to prove...

• Sep 19th 2009, 10:15 PM
IrrationalPI3
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

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

• Sep 19th 2009, 10:51 PM
NonCommAlg
Quote:

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

$\displaystyle a_n = \sqrt{2 +\sqrt {2+ \sqrt{2+\cdots + \sqrt2}}}$
"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 $\displaystyle a_1=\sqrt{2}$ and $\displaystyle a_{n+1}=\sqrt{2+a_n}, \ n \geq 1.$ now an easy induction over $\displaystyle n$ shows that $\displaystyle 0<a_n<2.$ then it follows that the sequence $\displaystyle \{a_n \}$ is increasing because

$\displaystyle a_{n+1}=\sqrt{2 + a_n} > a_n$ is equivalent to $\displaystyle (a_n + 1)(a_n - 2) < 0,$ which is obviously true because $\displaystyle 0 < a_n < 2.$
• Sep 19th 2009, 10:53 PM
red_dog
$\displaystyle a_n=\sqrt{2+\sqrt{2+\ldots+\sqrt{2}}}$

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

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

Suppose that $\displaystyle a_{n-1}<a_n$

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

Then $\displaystyle a_n<a_{n+1}$

$\displaystyle a_1=\sqrt{2}<2$

Suppose that $\displaystyle a_n<2$.

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