You can show the first part by the definition of limit.
"show that if xn converges to l, then 1/n (x1 + x2 + ... + xn) converges to l. then show that xn = n(log(n+1)-log(n)) = -1/n log(n!) + log(n+1) goes to 1". ok, the first bit here i don't know how to show it because it is not with any particular sequence. it is just an abstract 'show that this rule is true' and i dont know how to do that. i just know that it is from my notes. any guiders? the second part, i don't know what actually needs to be shown because it is all there already. the question kind of shows that already.
This is known as the sequence of means.
It is proved in two cases.
First suppose that .
If we can see that
The first sum is a fixed partial sum and by diving by can be made small.
Each term is the second sum are small because .
Then for the case define and apply case #1.
i found an answer to this question but it relies on the epsilon-N definition. and uses that |x1-l|/n < epsilon/n, |x2-l|/n < epsilon/n, etc, BUT isn't this only true for n > N so, you cant say that for the low terms x1, x2 etc..?
Yes. If you can use the claim when L = 0 (I'll write capital L because lowercase l can be confused with 1), then follow thw second part of Plato's advice:
Look, stop panicking whenever you see a letter used in a way you have not seen before. What has not been covered: that a sum of n terms can be split into two parts: the sum of the first K terms and the sum of the rest n - K terms? If you don't understand Plato's reasoning, at least you could have posed a more detailed question, e.g., how exactly is made small.
Yes; that's why it was suggested to split the series into terms up to some K and the rest.
OK, speaking in more detail: if you are allowed to use the claim when L = 0 and you are given a sequence that converges to an L ≠ 0, then following Plato's advice, define . Use the theorem about the limit of the sum (or difference) of sequences. Apply the claim for L = 0 to . Expand the definition of to see what the arithmetic mean of , ..., is.
Here is a Soraban complete proof.
Suppose that and .
There is such that if then . This is from convergence.
Now let . There is such that
O.K. Suppose that and if we get
.
Now note that if it follows that so that .
Moreover, because .
Now if you cannot finish, then you cannot be helped.