Does Cauchy's Theorem Confirm That Arithmetic Means of Null Sequences Are Null?

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Cauchy's Theorem asserts that the arithmetic means of a null sequence also form a null sequence. The proof involves selecting a natural number m based on a given ε, ensuring that for sufficiently large n, the terms of the sequence remain small. The discussion highlights confusion regarding the fixed nature of the numerator in the proof, clarifying that while m varies with ε, it is chosen to maintain certain properties that ensure convergence. The importance of making n large to minimize the fraction in the proof is emphasized, reinforcing the concept of limits in relation to ε. Overall, the conversation centers on understanding the nuances of the proof and the role of ε in determining convergence.
elliti123
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Hello , i was just wondering if anyone could clarify one thing in this proof (its from Konrad Knopp book on infinite series) : If (x0,x1,...) is a null sequence, then the arithmetic means

xn'= x0+x1+x2+...+x/n+1 (n=1,2,3,...)

also forms a null sequence.

Proof: If ε >0 is given, then m can be so chosen, that for every n > m we have |xn| < ε/2 . For these n's, we have
|xn'| ≤ |x1+x2+x3+...+xm| / n+1 +(ε/2) (n-m /n+1)

since the numerator of the first fraction on the right hand side now contains a fixed number, we can further determine n0, so that for n > n0 that fraction remains < ε/2. But then, for every n > n0 , we have |xn'| < ε and our theorem is proved.

My question is: the chosen m in the proof as far as i know is a natural number changing according to what epsilon we give it so for example if the chosen m is 3 it might work for a particular ε but might not for another ε less than the other ε we have chosen first . So i have come to a conclusion that m or n0 that every n should be more than so the sequence converges to a real number is a function of epsilon therefore it changes whenever epsilon does. Now how exactly is the numerator they describe in the proof a fixed number?
Since the m changes whenever ε does then it is logical to infer that the summation of those terms would obviously change. And would you please explain the last part of the proof after the inequality i seem to have some vivid idea but i don't think i still get the last part. Thanks.
 
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##\epsilon## is given, you choose a different m and n each time.
 
Oh sorry, the one thing you wanted to ask is how the numerator on the left is a fixed number. It isn't but m determines it. M is different each time. So that numerator is different each time. But you choose it to have a certain property, that everything else in the sequence is small. Then you make n large to make that fraction small. Then everything is small.
 
verty said:
Oh sorry, the one thing you wanted to ask is how the numerator on the left is a fixed number. It isn't but m determines it. M is different each time. So that numerator is different each time. But you choose it to have a certain property, that everything else in the sequence is small. Then you make n large to make that fraction small. Then everything is small.
Haha there is a lot of small going on there.Anyways jokes aside so in this kind of proofs , i mean in general for epsilon proofs you actually do consider the epsilon you choose or give or even the epsilon itself to be a "fixed" positive number right?
 
What Limit is all about, the value we found after evaluating limit what this value actually shows.
 
There are probably loads of proofs of this online, but I do not want to cheat. Here is my attempt: Convexity says that $$f(\lambda a + (1-\lambda)b) \leq \lambda f(a) + (1-\lambda) f(b)$$ $$f(b + \lambda(a-b)) \leq f(b) + \lambda (f(a) - f(b))$$ We know from the intermediate value theorem that there exists a ##c \in (b,a)## such that $$\frac{f(a) - f(b)}{a-b} = f'(c).$$ Hence $$f(b + \lambda(a-b)) \leq f(b) + \lambda (a - b) f'(c))$$ $$\frac{f(b + \lambda(a-b)) - f(b)}{\lambda(a-b)}...

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