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

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SUMMARY

Cauchy's Theorem confirms that the arithmetic means of a null sequence, defined as \( x_n' = \frac{x_0 + x_1 + ... + x_n}{n+1} \), also form a null sequence. The proof demonstrates that for any given \( \epsilon > 0 \), a natural number \( m \) can be chosen such that for all \( n > m \), the absolute value \( |x_n| < \frac{\epsilon}{2} \). This leads to the conclusion that \( |x_n'| < \epsilon \) for sufficiently large \( n \), thus proving the theorem. The discussion clarifies that while \( m \) varies with \( \epsilon \), the numerator of the arithmetic mean is determined by the fixed number of terms summed, which remains consistent in the context of the proof.

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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.
 

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