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What I did was write out the sequence (Xn) to notice that (Xn+1) (the next element in the sequence) is equal to 1 + 1/Xn (this was not intuitive to me). Knowing that both (Xn) and (Xn+1) equal the same limit you can set them equal to each other like in example 3.4.3, then solve. | What I did was write out the sequence (Xn) to notice that (Xn+1) (the next element in the sequence) is equal to 1 + 1/Xn (this was not intuitive to me). Knowing that both (Xn) and (Xn+1) equal the same limit you can set them equal to each other like in example 3.4.3, then solve. | ||
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| + | More detail about this solution method can also be found at http://www.mathacademy.com/pr/prime/articles/fibonac/index.asp | ||
| + | M.N. | ||
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Revision as of 16:27, 10 March 2010
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In #5, I'm having trouble figuring out how to prove that sequence $ Z $ is bounded in order to use Bolzano-Weierstrass Theorem and Theorem 3.4.9 to prove the necessary and sufficient conditions. I know that since both sequences $ X $ and $ Y $ are convergent that they are bounded, but I can't quite figure out how to use this information to prove that $ Z $ is bounded.
In #3, what will be a good starting point? I am having difficulty proceeding through the problem.
--Rrichmo 20:15, 10 March 2010 (UTC)
What I did was write out the sequence (Xn) to notice that (Xn+1) (the next element in the sequence) is equal to 1 + 1/Xn (this was not intuitive to me). Knowing that both (Xn) and (Xn+1) equal the same limit you can set them equal to each other like in example 3.4.3, then solve.
More detail about this solution method can also be found at http://www.mathacademy.com/pr/prime/articles/fibonac/index.asp
M.N.
