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 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
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 .
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.