Is the sequence Convergent?

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February 28th 2013, 03:02 AM
weijing85

Is the sequence Convergent?

Hi, my answer is True. Am i correct?

A sequence {a_n} satisfies a_n+1= 2a_n/(1+a_n) for n=1,2,....
If a_n=2, the sequence is monotonic decreasing and bounded, so it is convergent.

(a) True
(b) False

Thanks for checking (:
February 28th 2013, 05:02 AM
Esteban

Re: Is the sequence Convergent?

Quote:

Originally Posted by weijing85

Hi, my answer is True. Am i correct?

A sequence {a_n} satisfies a_n+1= 2a_n/(1+a_n) for n=1,2,....
If a_n=2, the sequence is monotonic decreasing and bounded, so it is convergent.

(a) True
(b) False

Thanks for checking http://mathhelpforum.com/number-theory/213981-sequence-convergent.html#post772672\" rel=\"nofollow\">

A sequence {a_n} satisfies a_n+1= 2a_n/(1+a_n) for n=1,2,.... If a_n=2, the sequence is monotonic decreasing and bounded, so it is convergent.
(a) True
(b) False

I suspect there is a typo in the post,
Is it not $a_1=2~?$ If so, use induction to show $a_n>a_{n-1}>1$ .

Hint: note that if $a>1$ then $a<a^2$ .

To find the limit solve $L=\frac{2L}{1+L}$ .
February 28th 2013, 01:58 PM
weijing85

Re: Is the sequence Convergent?

Thanks for the correction. Is a1 =2.
March 1st 2013, 06:25 PM
Nehushtan

Re: Is the sequence convergent?

Alternatively you can prove by induction that $a_n=\frac{2^n}{2^n-1}$ and take it from there. (The formula is easily guessed after trying a few values of $n$ .) (Smile)
March 2nd 2013, 12:50 AM
weijing85

Re: Is the sequence Convergent?

So the answer is true? :)
March 2nd 2013, 07:27 AM
Nehushtan

Re: Is the sequence convergent?

Well, that’s for you to find out, isn’t it? (Wink) I mean, it may be that you find the formula $a_n=\frac{2^n}{2^n-1}$ easier to work with; if not, you can always stick to the original recursive formula. (Cool)