# Thread: Is the sequence Convergent?

1. ## 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 (:

2. ## Re: Is the sequence Convergent?

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 (:
that's what exercise says? because, if $a_n=2$ it's a constant sequence, then it's trivially convergent

3. ## Re: Is the sequence Convergent?

Originally Posted by weijing85
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 $L=\frac{2L}{1+L}$.

4. ## Re: Is the sequence Convergent?

Thanks for the correction. Is a1 =2.

5. ## 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 $n$.)

6. ## Re: Is the sequence Convergent?

Well, that’s for you to find out, isn’t it? 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.