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Math Help - Urgent Question: Convergens of series

  1. #1
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    Urgent Question: Convergens of series

    Could somebody here please verify my proof?


    Given the two sequences \sum_{n=1}^{\infty} c_{n} and \sum_{n=1}^{\infty} d_{n} which is defined as

    d_{n} = \frac{c_1 + c_2 + \ldots + c_n}{n}


    1) show that if c_{n} \rightarrow 0, for n \rightarrow \infty is true. Then d_{n} \rightarrow 0 for n \rightarrow \infty


    2) show that if c_{n} \rightarrow c, for n \rightarrow \infty. Then d_{n} \rightarrow c for n \rightarrow \infty


    Solution 1)

    I choose an N' \in \mathbb{N} which for all m, m \geq N' implies that |c_m|< \frac{\epsilon}{2}.

    I then choose a random n > N' there is an N'' such that for all

    from which is follows that

    d_{n} = \frac{c_1 + c_2 + \ldots + c_N'}{n} + \frac{c_{N'+1}+ c_2 + \ldots + c_n}{n}

    Then by the triangle inequality

    \begin{array}{cccc}|d_n| &\leq& |\frac{c_1 + c_2 + \ldots + c_N'}{n}| + |\frac{c_{N'+1}+ c_2 + \ldots + c_n}{n}| \\ <br />
|d_n| &\leq& \frac{|c_1 + c_2 + \ldots + c_N'|}{n} + \frac{|c_{N'+1}|+ |c_2| + \ldots + |c_n|}{n} \end{array} \\<br />
|d_n| &\leq& \frac{|c_1 + c_2 + \ldots + c_N'|}{n} + \frac{(N-N')\frac{\epsilon}{2}}{2} \\ |d_n| &\leq&  \frac{|c_1 + c_2 + \ldots + c_N'|}{n} + \epsilon \end{array}\end{array}<br />

    there is an N'' such that for all n \geq N'', thus |d_n| < \epsilon

    Solution 2)

    y_n:=c_n-c \rightarrow 0 \mathrm{ as }n \rightarrow \infty \mathrm{ and thus } d_n-c=\frac{(c_1-c)+\cdots +(c_n-c)}{n}=\frac{y_1+\cdots +y_n}{n}
    will tend to 0 by part (a).


    I have let the two last tex modules stand because I cannot get them to work with "math", "/math"

    Best regards
    Last edited by Billy2007; December 13th 2006 at 11:44 PM.
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  2. #2
    Forum Admin topsquark's Avatar
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    This would look a lot nicer if you edited it and changed all the "tex" "/tex" labels to "math" "/math"

    -Dan
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