a1=1

a_n+1=a_n*1/2

how to prove that

a_n+1 -a_n<0

?

i tried by induction

a1>0

suppose a_n>0

a_n+1 -a_n>0

-a_n*1/2<0

so its monotonicly desendant correct?

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- Mar 4th 2011, 01:04 AMtransgalacticprooving monotonicity mmn 11
a1=1

a_n+1=a_n*1/2

how to prove that

a_n+1 -a_n<0

?

i tried by induction

a1>0

suppose a_n>0

a_n+1 -a_n>0

-a_n*1/2<0

so its monotonicly desendant correct? - Mar 4th 2011, 01:45 AMFernandoRevilla
Although it is very easy, you also need to prove that for all .

- Mar 4th 2011, 01:52 AMchisigma
It is a little 'surpising' the fact that You have a total of 973 posts and none of them is written in Latex!... the question is not minor because what You have written can cause misundestanding!...

If the 'recursive relation' is...

, (1)

... then is and the sequence is neither increasing nor decreasing...

If the 'recursive relation' is...

, (2)

... then is...

(3)

... and the (3) is negative so that the sequence is decreasing and it finishes at the n for which is ...

There are some other 'possible interpretations' but in any case the 'advatages' of the use of Latex are fully evident!...

Kind regards

- Mar 4th 2011, 02:32 AMFernandoRevilla
- Mar 4th 2011, 03:00 AMHallsofIvy
I would have interpreted that as !

To prove, by induction that this is a strictly decreasing sequence, that is, that for all n, note that and that so the statement is true for n=1.

Assume that for some k. Then but since , . That is, and we are done.

If it is, in fact, , then the sequence is neither increasing nor decreasing- for all n!