The discussion revolves around proving that if a sequence of real numbers {an} converges to L, then the average of the first n terms, [(a1 + a2 + ... + an)/n], also converges to L as n approaches infinity. The proof utilizes the epsilon-delta definition of limits, specifically that for any ε > 0, there exists an integer N such that for all n ≥ N, |an - L| < ε/4. The triangle inequality is employed to separate the average into two manageable terms, allowing for the demonstration that both terms can be made arbitrarily small, thus establishing the desired limit.
PREREQUISITESStudents of real analysis, mathematicians focusing on convergence proofs, and anyone interested in understanding the foundational concepts of limits and sequences.
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?