Prove: Advanced Calc Proof: ((a(n))+(a(n+1)))/2 Converges to A

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Homework Statement


If ((a(n))+(a(n+1)))/2 converges to A, then a(n) converges to A. Either prove or disprove this conjecture.



Homework Equations


Normal convergence proof



The Attempt at a Solution


I will prove that ((a(n))+(a(n+1)))/2 converges to A such that for every (epsilon)>0 there exists a positive integer N such that for every n>N abs(a(n)-A)<(epsilon).

Consider (epsilon)> 0 arbitrary.
Because a(n) and a(n+1) will converge to the same number...

That second part is what I am stuck at. I am really good at proving things converge this way with sequences defined but struggle in the abstracts. I struggle with the second and fourth lines of the proof, as the third is just consider n>N arbitrary.
 
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Consider the sequence 1, -1, 1, -1, ...that is a(n)=1 if n is odd, a(n)=-1 if n is even. What is the limit of (a(n)+a(n+1))/2?ehild
 
Wow, yeah. And I just used that to disprove the question before that. Thank you!

Another related question on proofs:

One is false and one is true (Prove the true, disprove the false)

Conjecture A: If a(n) is a sequence of real numbers then for every positive integer n there exists an M such that abs(a(n)) <= M.

Conjecture B: If a(n) is a sequence of real numbers there exists an M such that every positive integer n abs(a(n)) <= M.

I have been staring at it and I should know it but am struggling to figure it out.
 
bloynoys said:
Conjecture A: If a(n) is a sequence of real numbers then for every positive integer n there exists an M such that abs(a(n)) <= M.

Conjecture B: If a(n) is a sequence of real numbers there exists an M such that every positive integer n abs(a(n)) <= M.
I think you miss something from sentence B. abs(a(n)) <= M has to be true for every positive integer n.

Think what do the sentences mean? Imagine a simple sequence, 0.1, 0.2, 0.3,... for example. Is A true? Choose an n, for example n=1000. Can you find an M so as |an|<=M?

Can you find an M so as |a(n)|<=M for every n?

ehild
 
Thank you! That is what so frustrating about an entirely proof based class at this level. There is so many ways to look at these things and you understand it in class but when you get home to do questions that are slightly different you just sit there, as I know the stuff, but helps to have someone saying consider this... and then connecting the dots.
 
sometimes it helps to know what is really being said.

conjecture A states that every term of a real sequence is a finite real number.

conjecture B states that every real sequence is bounded.

put this way, it should be clear which one is true, and which one is false.

(and, on a more basic level, why the order of "for all" and "there exists" can't simply be reversed).
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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