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Math Help - limit of average of terms

  1. #1
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    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.
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  2. #2
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    Re: limit of average of terms

    You can show the first part by the definition of limit.
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    Re: limit of average of terms

    Quote Originally Posted by learning View Post
    "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.
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    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
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    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..?
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    Re: limit of average of terms

    Quote Originally Posted by learning View Post
    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 View Post
    Then for the case L\ne 0 define y_n=L-x_n and apply case #1.
    Quote Originally Posted by learning View Post
    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 View Post
    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.
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  7. #7
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    Re: limit of average of terms

    Quote Originally Posted by learning View Post
    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.
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    Re: limit of average of terms

    Hi, I just registered to say thanks. You guys really are incredible.
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