# prooving monotonicity mmn 11

• March 4th 2011, 01:04 AM
transgalactic
prooving 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?
• March 4th 2011, 01:45 AM
FernandoRevilla
Although it is very easy, you also need to prove that $a_n>0$ for all $n$ .
• March 4th 2011, 01:52 AM
chisigma
Quote:

Originally Posted by transgalactic
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?

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...

$\displaystyle a_{n+1}= \sqrt{a_{n}}$ , $a_{1}=1$ (1)

... then is $\forall n\ \ a_{n}=1$ and the sequence is neither increasing nor decreasing...

If the 'recursive relation' is...

$\displaystyle a_{n}+1= \sqrt{a_{n}}$ , $a_{1}=1$ (2)

... then is...

$\displaystyle \Delta_{n} = a_{n-1}-a_{n} = \sqrt{a_{n}} - a_{n}- 1$ (3)

... and $\forall a_{n}>0$ the (3) is negative so that the sequence is decreasing and it finishes at the n for which is $a_{n}<0$...

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

Kind regards

$\chi$ $\sigma$
• March 4th 2011, 02:32 AM
FernandoRevilla
Quote:

Originally Posted by chisigma
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!...

I agree.
• March 4th 2011, 03:00 AM
HallsofIvy
I would have interpreted that as $a_{n+1}= \frac{a_n}{2}$!

To prove, by induction that this is a strictly decreasing sequence, that is, that $a_{n+1}< a_n$ for all n, note that $a_1= 1$ and that $a_2= \frac{1}{2}$ so the statement is true for n=1.

Assume that $a_{k+1}< a_k$ for some k. Then $a_{k+2}= \frac{a_{k+1}}{2}$ but since $a_{k+1}< a_k$, $\frac{a_{k+1}}{2}< \frac{a_k}{2}= a_{k+1}$. That is, $a_{k+2}< a_{k+1}$ and we are done.

If it is, in fact, $a_{n+1}= \sqrt{a_n}$, then the sequence is neither increasing nor decreasing- $a_n= 1$ for all n!