Question

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~

Answer 1

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
\[x=2\]

Answer 2

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 \[a_n+1/a_n > 1\]

Answer 3

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
\[a_n=\sqrt{2a_{n-1}}\leq \sqrt{2\times 2} \text{ by induction } <\sqrt{4}=2\]

Answer 4

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

Answer 5

you get

\[\frac{2(a_n-a_{n-1})}{\sqrt{2a_n}+\sqrt{2a_{n-1}}}\]

Answer 6

induction will show that this is greater than zero

Answer 7

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

Answer 8

yw