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.
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..?
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.
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.