The discussion revolves around proving that the average of a converging sequence of real numbers also converges to the same limit. The original poster presents a sequence {an} that converges to L as n approaches infinity and seeks to demonstrate that the average (a1 + a2 + ... + an) / n also approaches L.
Participants are actively engaging with the problem, raising questions about specific steps in the proof and discussing the implications of boundedness due to convergence. Some guidance has been offered regarding the use of the triangle inequality and the need to establish bounds for different terms in the expression.
There is an ongoing discussion about the necessity of finding specific integers N and N1 that satisfy the conditions of the proof, as well as the challenge of relating the average to the limit in a rigorous manner. Participants are considering how the terms of the sequence influence the choice of these integers.
kingwinner said:I think this would work.
Take N=max{N1,N2}
where N1 is as you defined
and N2 = the smallest integer greater than or equal to 2 |(a1-L)+...+(aN-L)| / ε.
If n > N, then |[(a1+a2+...+an)/n] - L|< ε.
In our back-and-forth we may have called N different things; that's why I summarized it in post #29. We don't need an N3.However, according to what you said in post #7, we should take N to be the max of THREE things, i.e. N=max{N1,N2,N3}. Why? What is N3? I don't understand where this third restriction comes from.
I believe |[(aN+1 - L) +...+ (an-1-L) +(an-L)] / n| <= (1/n)[|aN+1 - L| + ...+ |an-1-L| + |an-L|] < (1/n)(n-N)ε/4 < ε/4 is always true, isn't it? So what is N3?