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Math Help - Limit Point Question

  1. #1
    Forum Admin topsquark's Avatar
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    Limit Point Question

    Suppose a series z_n \to \alpha as n \to \infty.

    (That is to say ( z_n ) has a limit point \alpha at infinity.)

    Prove that
    \lim_{n \to \infty} \frac{z_1 + z_2 + ~...~ + z_n}{n} = \alpha

    I have found a number of series where I can prove it case by case, but am having some trouble generalizing. In addition, if the series z_n = 1 - \frac{1}{n} I am having an embarrassing time showing the result at all.

    Another question: Does this theorem have an actual use? I can't think of one.

    (FYI: This problem is actually designed for the series to be complex. I figure if I can find out how to prove it for real series then I can generalize it to complex numbers myself.)

    Thanks!
    -Dan
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  2. #2
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    Quote Originally Posted by topsquark View Post
    Suppose a series z_n \to \alpha as n \to \infty.

    (That is to say ( z_n ) has a limit point \alpha at infinity.)

    Prove that
    \lim_{n \to \infty} \frac{z_1 + z_2 + ~...~ + z_n}{n} = \alpha

    I have found a number of series where I can prove it case by case, but am having some trouble generalizing. In addition, if the series z_n = 1 - \frac{1}{n} I am having an embarrassing time showing the result at all.

    [snip]
    If z_n = 1 - \frac{1}{n} then

    \frac{z_1 + z_2 + ~...~ + z_n}{n} = \frac{ \left( 1 - \frac{1}{1} \right) + \left( 1 - \frac{1}{2} \right) + \left( 1 - \frac{1}{3} \right) + ~...~ + \left( 1 - \frac{1}{n} \right)}{n}


    = \frac{n - \sum_{k=1}^{n} \frac{1}{k}}{n}


    = 1 - \frac{\sum_{k=1}^{n} \frac{1}{k}}{n}.


    \lim_{n \rightarrow \infty} \frac{\sum_{k=1}^{n} \frac{1}{k}}{n} = 0 since the harmonic series diverges to infinity a heck of a lot slower than n does (I know that's not a solid technical argument but I'm sure it can be made so easily ..... is the glib 'easily' business the sticking point?)
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  3. #3
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    Quote Originally Posted by topsquark View Post
    Suppose a series z_n \to \alpha as n \to \infty.

    (That is to say ( z_n ) has a limit point \alpha at infinity.)

    Prove that
    \lim_{n \to \infty} \frac{z_1 + z_2 + ~...~ + z_n}{n} = \alpha

    I have found a number of series where I can prove it case by case, but am having some trouble generalizing.
    [snip]
    This seems way too glib to be a correct proof:

    If z_n \rightarrow \alpha as n \to \infty, then z_1 \rightarrow \alpha, z_2 \rightarrow \alpha, .....

    Then \lim_{n \to \infty} \frac{z_1 + z_2 + ~...~ + z_n}{n} = \lim_{n \to \infty} \frac{\alpha + \alpha + ~...~ + \alpha}{n} = \lim_{n \to \infty} \frac{n \alpha}{n} = \alpha ....

    But like I said ....
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    Quote Originally Posted by topsquark View Post
    Suppose a series z_n \to \alpha as n \to \infty.

    (That is to say ( z_n ) has a limit point \alpha at infinity.)
    Do you really mean series or do you mean sequence?
    One does not usually say that a series has a limit point.
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    I proved this for my analysis class last year . It usues real numbers, but you can easily generalize it. Here.
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    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Plato View Post
    Do you really mean series or do you mean sequence?
    One does not usually say that a series has a limit point.
    Yes, you are correct. I should have said sequence. (What can I say, it was a late night last night!)

    -Dan
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    Forum Admin topsquark's Avatar
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    Quote Originally Posted by mr fantastic View Post
    \lim_{n \to \infty} \frac{z_1 + z_2 + ~...~ + z_n}{n} = \lim_{n \to \infty} \frac{\alpha + \alpha + ~...~ + \alpha}{n} = \lim_{n \to \infty} \frac{n \alpha}{n} = \alpha ....
    Thank you for the effort, but this statement is incorrect. Perhaps the mistake in terminology Plato pointed out is a problem. So let me restate the original problem in more correct terms:

    Suppose a sequence ( z_n ) has the property z_n \to \alpha as n \to \infty.

    (That is to say ( z_n ) has a limit point \alpha at infinity.)

    Prove that
    \lim_{n \to \infty} \frac{z_1 + z_2 + ~...~ + z_n}{n} = \alpha

    Since we are talking about a sequence here we cannot say that z_n \to \alpha for all n, we are saying that the sequence approaches a limit point \alpha, as n approaches infinity. That is to say there is only one sequence, not an infinite number of sequences tending toward the same limit.

    -Dan
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    Forum Admin topsquark's Avatar
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    Quote Originally Posted by ThePerfectHacker View Post
    I proved this for my analysis class last year . It usues real numbers, but you can easily generalize it. Here.
    I am somewhat confused almost from the beginning by the line
    | s _n - s | = \left | \frac{s_1 + ~...~ + s_n - s \cdot n}{n} \right |

    Is this again a case of my mistaken wording? If you define s_n to be the nth partial sum of a series, this statement is quite understandable and falls out naturally. (Hopefully by now it is clear that this is not what I meant.)

    -Dan
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    Quote Originally Posted by topsquark View Post
    Suppose a series z_n \to \alpha as n \to \infty.

    (That is to say ( z_n ) has a limit point \alpha at infinity.)

    Prove that
    \lim_{n \to \infty} \frac{z_1 + z_2 + ~...~ + z_n}{n} = \alpha
    There's a neat proof of this by CaptainBlack here for the case where α=0. You can easily deduce the general case from this by applying it to the sequence (z_n-\alpha).

    Quote Originally Posted by topsquark View Post
    Another question: Does this theorem have an actual use? I can't think of one.
    This is a case of CesÓro summation, which is an essential tool in Fourier theory. The main result about convergence of Fourier series (FejÚr's theorem) is proved by using CesÓro summation.

    So yes, the fact that ordinary convergence implies CesÓro convergence is certainly useful.
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    Topsquark try proving this similar result: Let x_n>0 and \lim ~ x_n = x. Prove that the geometric means (x_1...x_k)^{1/k} \to x.
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  11. #11
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    Quote Originally Posted by topsquark View Post
    I am somewhat confused almost from the beginning by the line
    | s _n - s | = \left | \frac{s_1 + ~...~ + s_n - s \cdot n}{n} \right |

    Is this again a case of my mistaken wording? If you define s_n to be the nth partial sum of a series, this statement is quite understandable and falls out naturally. (Hopefully by now it is clear that this is not what I meant.)

    -Dan
    I am using \sigma_n not s_n. Where \sigma_n is the sequence of averages.
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