In class we were given the following sequence-(adsbygoogle = window.adsbygoogle || []).push({});

(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)

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# Proving that a sequence is confined to a certain interval

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