Sequence of real numbers | Proof of convergence

by kingwinner
Tags: convergence, numbers, proof, real, sequence
 P: 1,270 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]
 P: 1,270 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!!!
 P: 1,635 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 0a_0+a_1[/tex] If we continue in this fashion, after n-2 steps we would get something like: $$a_{n+2}=\sqrt{a_{n+1}}+\sqrt{a_n}>a_0+a_1+...+a_n>n*min\{a_0,a_1,...,a_ n\}=n*a$$ So, now you see that if we let n>N=1/a we get our result. where a=min{a_o,...,a_n} cheers!
P: 1,270

Sequence of real numbers | Proof of convergence

Thanks.
Using part a, I proved part b.