Proving that a sequence is confined to a certain interval

[Claire84]

In class we were given the following sequence-

(0,1 1/2, 3/4, 1 1/8, 15/16, 1 1/32 ...)

So it's alternating about zero. We're told then that given any number a>1 it s true to say that the sequence (xn) is eventually confined t [0,a]

We now have to prove this...

Inworked out the formula for the sequence t be 1- (-1/2)^(n-1) but I'm not sure if that was necessary or not.

Then I set about getting the poof by working backwards to get an idea of what it was going to be like, but I'm confuse because it's an alternating sequence and we haven't done any like that yet. I asked one o the phd guys about it and they looked at it for half an hour and couldn't do it, but it's only a level 1 problem so can someone else give some feedback on it? Here's what I've come up with so far for my idea for the proof...

Since we have n>no

1- (-1/2)n-1 > 1-(-1/2)^(no-1) for even powers...

so we have a> 1- (1/2)^(n-1)

(-1/2)^(n-1)>1-a

since it's an even power you could then have-

1/2^(n-1)> 1-a

2/(2^n)>1-a

so you could then start the proof with a number (1-a)/2>0

Hoqwever, I don't know what to do for uneven powers because then wouldn't 1-(-1/2)^(n-1) < 1- (-1/2)^(no -1)

 

[matt grime]

You're on the right track.

Let a_n be the sequence 1-(-1/2)^n as given

we must show that for all a>1, a_n<a for all n sufficiently large.

Given a, let N be chosen such that 2^(-N)<a-1, which can be done (by eg archimedes' axiom)

|a_n| <= 1+(1/2)^n for all n (ie we can ignore the minus sign) and if n>N then |a_n| < 1+(1/2)^N < a

 

[Claire84]

|a_n| <= 1+(1/2)^n for all n (ie we can ignore the minus sign)

is this because this would have bee like the odd power and therefore the 2 minuses cancel out? And then you don't really need to think about the even power thing I dscribed because it's obviously going to be less than one? Would you still need to go through the even power situation? Sorry, this is all a bit new to me but I'm glad I actually managed to do something relatively logical in my first post.

 

[matt grime]

yep, (-x)^n = -(x^n) if n is odd, so the signs cancel. you at least need to state that the n even case is clearly always less than 1

 

[Claire84]

Aha, thanks for that! Suddenly feeling very brainy! *cough* ell a least it shows I'm enthusiatsic that I'm doing Maths on St.Patrick's day!