[note: also under discussion in s.o.s. math board]
I don't see how to work out the epsilon-limit proof.
At the end, we have to show that:
for all ε>0, there exists an integer M such that
n≥M => |[(a1+a2+...+an)/n] - L|< ε.
But how can we find that M?
(even for the case L=0 I am facing the same trouble...I can't find a way to link |[(a1+a2+...+an)/n] - L| with |a_n -L|, so I don't know how to find M)
Any help is appreciated!
I would LOVE to have it, but unfortunately I don't have the fortune to have such a treasury experience...I'm learning it from my textbooks.
But could you kindly point out which part I was wrong that leads you to make the above comment? I don't think I've said anything conceptually wrong yet...
We have to show that (by definition of convergence):
for all ε>0, there exists an integer M such that
n≥M => |[(a1+a2+...+an)/n] - L|< ε.
and just like every epsilon-limit proof of this type, we have to actually find a specific M that works, right? And to do this, I believe we need to link |[(a1+a2+...+an)/n] - L| with |a_n -L|. And right now, I am stuck on this step. I know that the triangle inequality may help, but I am not sure how to apply it in this case.
Could someone help me, please?
Any help is much appreciated!
OK, after using the triangle inequality, I will have n terms.
|(a_1 -L)/n|, |(a_2 -L)/n|,...,|(a_(n-1) -L)/n|, |(a_n -L)/n|
How can I bound each of these?
My idea is to set each of these to be less than ε/? (I don't know what to use for the ? part) and solve the inequality for n, but since it contains terms like a_(n-1) and a_n in the numerator, I don't think I can solve for n?
I have no clue how to continue, can you please show me how?
Thanks!
By the way, I obliged you, but i wouldn't really use the triangle inequality here, well, not like this at least. finding something to bound each of these expressions seems like too much work, or at least odd. pick your to be one of the subscripts in . Bound the sum of all of everything below it (that is, the expressions for which ), and each term above it could be bounded by some . See if you can make something like that work.