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How do I prove by limits that this sequence is less than 2 for all values of n?
Last edited by simplyjasper (2009-11-16 06:20:03)
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Use mathematical induction
Assume
Then
Hence, by induction we have
Last edited by betterthangauss (2009-11-16 06:55:51)
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Here's my solution, but I don't know if it involves limits in the way you were thinking
Last edited by Identity (2009-11-16 06:55:12)
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Hi simplyjasper,
You can prove that the sequence is less than 2 by mathematical induction together with finding the limit.
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Hi simplyjasper;
Here is a proof by contradiction:
Suppose for some m that a(m) >= 2 then a(m+1)>=2:
If we square both sides of the recurrence we get:
a(m+1)^2 = a(m) +2
We run the recursion backwards to get:
a(m) = a(m+1)^2 - 2
Since a(m+1) >=2 then a(m) >=2.
We can now continue all the way back to a(1) >=2.
But this is a contradiction so a(n+1) = √(a(n) +2) is never >= 2.
Last edited by bobbym (2009-11-16 23:03:44)
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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