# limit of average of terms

• Mar 27th 2013, 09:39 AM
learning
limit of average of terms
"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.
• Mar 27th 2013, 10:09 AM
emakarov
Re: limit of average of terms
You can show the first part by the definition of limit.
• Mar 27th 2013, 10:11 AM
Plato
Re: limit of average of terms
Quote:

Originally Posted by learning
"show that if xn converges to l, then 1/n (x1 + x2 + ... + xn) converges to l.

This is known as the sequence of means.
It is proved in two cases.
First suppose that $L=0$.
If $1\le K\le n$ we can see that $\left| {\frac{1}{n}\sum\limits_{k = 1}^n {{x_k}} } \right| \le \frac{1}{n}\left| {\sum\limits_{k = 1}^K {{x_k}} } \right| + \frac{1}{n}\left| {\sum\limits_{k = K + 1}^n {{x_k}} } \right|$

The first sum is a fixed partial sum and by diving by $n$ can be made small.

Each term is the second sum are small because $L=0$.

Then for the case $L\ne 0$ define $y_n=L-x_n$ and apply case #1.
• Mar 27th 2013, 10:51 AM
learning
Re: limit of average of terms
it is stated that it is true for 0. so i allowed can use this information without proving it.

does that affect what i should do. i dont know about any capital K .that has never been covered
• Mar 27th 2013, 12:53 PM
learning
Re: limit of average of terms
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..?
• Mar 27th 2013, 01:25 PM
emakarov
Re: limit of average of terms
Quote:

Originally Posted by learning
it is stated that it is true for 0. so i allowed can use this information without proving it.

does that affect what i should do.

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:

Quote:

Originally Posted by Plato
Then for the case $L\ne 0$ define $y_n=L-x_n$ and apply case #1.

Quote:

Originally Posted by learning
i dont know about any capital K .that has never been covered

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 $\frac{1}{n}\left| {\sum\limits_{k = 1}^K {{x_k}} } \right|$ is made small.

Quote:

Originally Posted by learning
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; 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 $y_n=L-x_n$. Use the theorem about the limit of the sum (or difference) of sequences. Apply the claim for L = 0 to $y_n$. Expand the definition of $y_n$ to see what the arithmetic mean of $y_1$, ..., $y_n$ is.
• Mar 27th 2013, 04:36 PM
Plato
Re: limit of average of terms
Quote:

Originally Posted by learning
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..?

Here is a Soraban complete proof.

Suppose that $(x_n)\to 0$ and $c>0$.
There is $J\in\mathbb{Z}^+$ such that if $n\ge J$ then $\left| {{x_n}} \right| < \frac{c}{2}$. This is from convergence.

Now let $A = \sum\limits_{k = 1}^J {\left| {{x_k}} \right|}$. There is $K\in\mathbb{Z}^+$ such that $\frac{A}{K} < \frac{c}{2}$

O.K. Suppose that $N=J+K$ and if $n\ge N$ we get
$\left| {\sum\limits_{k = 1}^n {\frac{{{x_k}}}{n}} } \right| \le \frac{1}{n}\sum\limits_{k = 1}^J {\left| {{x_k}} \right|} + \frac{1}{n}\sum\limits_{k = J + 1}^n {\left| {{x_k}} \right|}$.

Now note that if $n\ge N$ it follows that $n>K$ so that $\frac{1}{n}\sum\limits_{k = 1}^J {\left| {{x_k}} \right|} \le \frac{1}{K}A < \frac{c}{2}$.

Moreover, $\frac{1}{n}\sum\limits_{k = j + 1}^n {\left| {{x_k}} \right|} \le \frac{{n - J}}{n}\frac{c}{2} < \frac{c}{2}$ because $\frac{{n - J}}{n} = 1 - \frac{J}{n} < 1$.

Now if you cannot finish, then you cannot be helped.
• Apr 2nd 2013, 03:42 AM