Cauchy null sequence proof (proof check) i was just wondering if anyone could clarify one thing in this proof (its from Konrad Knopp book on infinite series) : If $(x_0,x_1,...)$ is a null sequence, then the arithmetic means
$$x_n'= {x_0+x_1+x_2+...+x_n \over n+1}\qquad (n=1,2,3,....)$$
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
$$|x_n'| ≤ {|x_1+x_2+x_3+...+x_m|\over n+1} + {ε\over 2} {n-m \over n+1}$$
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.

Yes, your proof is correct, but maybe you need to look at it from another perspective. Your objection, if I understand correctly, is the fact that $m$ itself depends upon $\epsilon$; furthermore, you claim, that since the numerator itself changes if you change $\epsilon$, the numerator ceases to be  a fixed number. That objection is correct, though irrelevant. In proving that something converges to something else, it's sufficient to show that the difference gets smaller than any positive $\epsilon$, which you have done. 
As for the last part of the proof, you have it right, since $\frac{\epsilon}{2} \frac{n-m}{n+1}$ is always less than $\frac{\epsilon}{2}$, it suffices to show that the left term becomes less than $\frac{\epsilon}{2}$. That happens because $\frac{1}{n}$ goes to zero.
