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 (:

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- Feb 28th 2013, 03:02 AMweijing85Is 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 (: - Feb 28th 2013, 05:02 AMEstebanRe: Is the sequence Convergent?
- Feb 28th 2013, 06:32 AMPlatoRe: Is the sequence Convergent?
- Feb 28th 2013, 01:58 PMweijing85Re: Is the sequence Convergent?
Thanks for the correction. Is a1 =2.

- Mar 1st 2013, 06:25 PMNehushtanRe: Is the sequence convergent?
Alternatively you can prove by induction that $\displaystyle a_n=\frac{2^n}{2^n-1}$ and take it from there. (The formula is easily guessed after trying a few values of $\displaystyle n$.) (Smile)

- Mar 2nd 2013, 12:50 AMweijing85Re: Is the sequence Convergent?
So the answer is true? :)

- Mar 2nd 2013, 07:27 AMNehushtanRe: 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 $\displaystyle a_n=\frac{2^n}{2^n-1}$ easier to work with; if not, you can always stick to the original recursive formula. (Cool)