Thread: 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 (:

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 (:

that's what exercise says? because, if it's a constant sequence, then it's trivially convergent

Originally Posted by weijing85

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 If so, use induction to show .

Hint: note that if then .

To find the limit solve .

Thanks for the correction. Is a1 =2.

Alternatively you can prove by induction that and take it from there. (The formula is easily guessed after trying a few values of .)

So the answer is true?

Well, that’s for you to find out, isn’t it? I mean, it may be that you find the formula easier to work with; if not, you can always stick to the original recursive formula.