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Math Help - Arithmetic Means

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    Arithmetic Means

    For any sequence s_n consider the arithmetic mean

    t_n = (s1 + s2 + ... + sn) / n

    Prove sn -> s implies tn -> s. Prove there are divergent sequences s_n which in this manner give rise to convergent sequences t_n.
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    MHF Contributor chisigma's Avatar
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    Re: Arithmetic Means

    Quote Originally Posted by veronicak5678 View Post
    For any sequence s_n consider the arithmetic mean

    t_n = (s1 + s2 + ... + sn) / n

    Prove sn -> s implies tn -> s. Prove there are divergent sequences s_n which in this manner give rise to convergent sequences t_n.
    If s_{n} \rightarrow s we can set each s_{k}= s + \delta_{k} where \delta_{k} \rightarrow 0. Now is...

    t_{n} =\sum_{k=1}^{n} \frac{s+\delta_{k}}{n} = s + \sum_{k=1}^{n} \frac{\delta_{k}}{n} (1)

    ... and because each term of the sum at the second term of (1) tends to 0 the sum itself tends to 0...

    An example of divergent series that produces a convergent sequence t_{n} is the armonic series. Ibn effect is...

    \sum_{k=1}^{n} \frac{1}{k} \sim \gamma + \ln n (2)

    ... where \gamma is the 'Euler's constant'. From (2) You immediately derive that t_{n} \rightarrow 0...

    Kind regards

    \chi \sigma
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  3. #3
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    Re: Arithmetic Means

    Quote Originally Posted by veronicak5678 View Post
    For any sequence s_n consider the arithmetic mean

    t_n = (s1 + s2 + ... + sn) / n

    Prove sn -> s implies tn -> s. Prove there are divergent sequences s_n which in this manner give rise to convergent sequences t_n.
    Given \varepsilon>0, there exists N such that |s_k-s|<\varepsilon whenever k > N. Write t_n as

    t_n-s = \frac{(s_1-s)+(s_2-s)+\ldots+(s_N-s)}n + \frac{(s_{N+1}-s)+(s_{N+2}-s)+\ldots+(s_n-s)}n ,

    then show that the first of those fractions tends to 0, and the second fraction stays less than \varepsilon, as n\to\infty.

    For the second part of the question, what about the sequence s_n=(-1)^n?
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