## anonymous 4 years ago how to prove the convergence of this sequence using the Bounded Monotone convergent theorem : a_1 = sqrt(2) ; a_(n+1) = sqrt(2*a_n) ; and what is its limit ? ... thanks~

1. anonymous

limit is the easy part you have $\lim_{n\rightarrow \infty} a_n=\lim_{n\rightarrow \infty}\sqrt{2a_n}$ if you call that limit say x, then you see that $x=\sqrt{2x}$meaning $x^2=2x$ or $x=2$

2. anonymous

is it possible to prove the fact the sequence is strictly increasing by doing the ratio test ? $a_n+1/a_n = \sqrt(2*a_n) /a_n = \sqrt(2/a_n) =....$ but the problem is i don't know how to continue from the ... part. ( to show that the original question is strictly increasing i need to show $a_n+1/a_n > 1$

3. anonymous

if you need bounded you need induction. you can in fact assert that $a_n<2$ for all n step 1 of the inductions is $a_1=\sqrt{2}<2$ and then assume it is true for all k < n and then $a_n=\sqrt{2a_{n-1}}\leq \sqrt{2\times 2} \text{ by induction } <\sqrt{4}=2$

4. anonymous

monotone i think again by induction, this time with $a_{n+1}-a_{n}=\sqrt{2a_n}-\sqrt{2a_{n-1}}$ then use $\sqrt{a}-\sqrt{b}=\frac{a-b}{\sqrt{a}+\sqrt{b}}$

5. anonymous

you get $\frac{2(a_n-a_{n-1})}{\sqrt{2a_n}+\sqrt{2a_{n-1}}}$

6. anonymous

induction will show that this is greater than zero

7. anonymous

hmm i see, yeah so this does prove a_n is increasing XD~! thank you sooooo much XD~!

8. anonymous

yw