sequences- covergent?

**genlovesmusic09** · December 6th 2009, 02:20 PM

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

I have no idea how to beginning this...

**pickslides** · December 6th 2009, 02:27 PM

Originally Posted by genlovesmusic09

${a_{n + 1}} = \sqrt {2{a_n}} <br />$
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 $a_n = 1$ it looks like it converges to 2.

**genlovesmusic09** · December 6th 2009, 02:32 PM

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

but ${a_n} = 1$ ?

**hjortur** · December 6th 2009, 03:08 PM

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

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

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

And

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

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

Now what happens if $a_n<2$ ??

Hope that helps.

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