# Is the sequence Convergent?

• February 28th 2013, 03:02 AM
weijing85
Is the sequence Convergent?
Hi, my answer is True. Am i correct?

A sequence {an} satisfies an+1= 2an/(1+an) for n=1,2,....
If an=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 {an} satisfies an+1= 2an/(1+an) for n=1,2,....
If an=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 {an} satisfies an+1= 2an/(1+an) for n=1,2,.... If an=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.

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)