Hello , i was just wondering if anyone could clarify one thing in this proof (its from Konrad Knopp book on infinite series) : If (x(adsbygoogle = window.adsbygoogle || []).push({}); _{0},x_{1},...) is a null sequence, then the arithmetic means

x_{n}'= x_{0}+x_{1}+x_{2}+...+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 |x_{n}| < ε/2 . For these n's, we have

|x_{n}'| ≤ |x_{1}+x_{2}+x_{3}+...+x_{m}| / 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 n_{0}, so that for n > n_{0}that fraction remains < ε/2. But then, for every n > n_{0}, we have |x_{n}'| < 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 n_{0}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.

**Physics Forums | Science Articles, Homework Help, Discussion**

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Cauchy's theorem on limits

Tags:

**Physics Forums | Science Articles, Homework Help, Discussion**