Thread: sequences- covergent?

$\displaystyle {a_{n + 1}} = \sqrt {2{a_n}}
$
Is this a convergent sequence and if so, what does it converge to?


I have no idea how to beginning this...

Originally Posted by genlovesmusic09

$\displaystyle {a_{n + 1}} = \sqrt {2{a_n}}
$
Is this a convergent sequence and if so, what does it converge to?


I have no idea how to beginning this...

What is the first term?



Graphing some terms for $\displaystyle a_n = 1$ it looks like it converges to 2.

I wasn't given a first term which is why I am having I problem

but $\displaystyle {a_n} = 1$?

It is easy to see that the series converges for any value of $\displaystyle a_1>0$
if $\displaystyle a_1=0$ then the series converges to zero.

Now lets assume that $\displaystyle a_n>2$ then it follows that:

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

And

$\displaystyle a_{n+1}=\sqrt{2a_n}>\sqrt{2\cdot2}=2$

So you see that if $\displaystyle a_n>2$ then: $\displaystyle 2<a_{n+1}<a_n$.

Now what happens if $\displaystyle a_n<2$ ??

Hope that helps.