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Sequence of real numbers | Proof of convergence |
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| Jan23-10, 11:40 PM | #1 |
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Sequence of real numbers | Proof of convergence
1. The problem statement, all variables and given/known data
2. Relevant equations N/A 3. The attempt at a solution Assuming the truth of part a, I proved part b. But now I have no idea how to prove parts a & c. Part a seems true intuitively. The sqaure root of a number between 0 and 1 is will be larger than that number, and if we take more and more square roots, it will get close to 1, and then if we add two numbers that are close to 1, it must be ≥1. But how can we write a FORMAL proof of it? How can we find/construct N and demonstrate exactly that there exists an N such that n≥N => an≥1? Can someone help me, please? Any help is much appreciated! [note: also under discussion in Math Links forum] |
| Jan25-10, 02:59 AM | #2 |
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For part a, I have some more idea...if a(n) and a(n+1) are positive then we can surely pick an m such that 1/(2m) is less than them both: just select a sufficiently large m, e.g. select m such that 2m ≥ 1/min{a(n),a(n+1)}.
But how can I find N such that n≥N implies a(n)≥1 ? Any help is much appreciated!!! |
| Jan25-10, 04:06 AM | #3 |
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Since you seem to have spent quite some time on the problem i will try to give you some hints, i hope i don't get another warning from pf moderators for offering too much help(solving >90% of the problem for the op) :(
This might not be the nicest proof in the world, but i think it works. As you have figured out the main problem is when 0<a_1<1 and 0<a_o<1. So we will deal with this case only, since others are trivial. Let: [tex]0<a_0<1,0<a_1<1[/tex] then: [tex]a_o<\sqrt{a_o}...and...a_1<\sqrt{a_1}[/tex] adding these together we get: [tex]a_3=\sqrt{a_0}+\sqrt{a_1}>a_0+a_1[/tex] If we continue in this fashion, after n-2 steps we would get something like: [tex]a_{n+2}=\sqrt{a_{n+1}}+\sqrt{a_n}>a_0+a_1+...+a_n>n*min\{a_0,a_1,...,a_ n\}=n*a[/tex] So, now you see that if we let n>N=1/a we get our result. where a=min{a_o,...,a_n} cheers! |
| Jan25-10, 05:21 PM | #4 |
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Sequence of real numbers | Proof of convergence
Thanks.
Using part a, I proved part b. Any hints about part c? |
| Jan26-10, 12:26 PM | #5 |
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For part c, I'm stuck with using the hint.
From part b, en+2 ≤ (en+1 + en)/3 for n≥N. In part c, I think I need to end up proving something like en ≤ (2/3)some exponent involving n max(eN,eN+1) If the RHS tends to 0, then by squeeze theorem en->0. But I'm not sure how to SET UP the iteration. From part b, en+2 ≤ (en+1 + en)/3 for n≥N. Is it also true that en+1 ≤ (en + en-1)/3? Why or why not? And how can I find that "some exponent involving n"? May someone help me, please? Thanks!!!! |
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