# Homework Help: Series and their limits

1. Dec 7, 2008

hi :]

a couple of questions:

1) Using epsilon and N, write in a formal manner the following statement:
L is not a the limit of the general series {an} when n goes from 1 to infinity.

2) prove the next sentence: if a series an is converging into a final limit L, then the arithmetic avareges of the series organs(terms?) are gathering into the same limit. meaning:
lim (an)[n->infinity] = L = = = > lim [n->infinity] (a1+a2+a3..+an) / n = L

excuse my english.. not my strongest side.
I realy wish I could write down my attempts to solve the question by they are all in hebrew and are too hard to translate since I'm not sure myself that I'm on the right path..

Thanks,
sharon.

2. Dec 7, 2008

### CompuChip

Recall the definition of limit:
$$\lim_{n \to \infty} a_n = L$$ means that $$\forall \epsilon > 0, \cdots$$ ?

Then for 1 negate that statement:
$$\lim_{n \to \infty} a_n \neq L$$ means that $$\neg(\forall \epsilon > 0, \cdots) \Leftrightarrow \exists \epsilon > 0, \cdots$$ ?

For 2, you will somehow need to estimate the arithmetic average (yes, they are called terms, although organs is a nice one as well ). That is, if you know that an comes arbitrarily close to L, then you want to show the same for (a1 + ... + an)/n.

3. Dec 7, 2008

thanks, but i didnt realy understood (1) ..

4. Dec 7, 2008

### CompuChip

OK, first step:
what is the definition of
$$\lim_{n \to \infty} a_n = L$$

5. Dec 7, 2008

the limit exists if for each ε > 0 there exists an R such that qqq |f(x) - L| < ε whenever x > R

so the limit does not exists when |f(x) - L| < ε whenever x < R ?

6. Dec 8, 2008

### CompuChip

Right.

No. The limit is not L, if it is not true that for each ε > 0 there exists an R such that |f(x) - L| < ε whenever x > R. In a first mathematics course you must have learned how to rewrite such a statement. Things like: if it is not true that all cows eat grass, then there must exist a cow who does not eat grass. In this case, your answer would start with: "the limit is not L, when there exists an ε > 0, ..."