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Math Help - Proving that (1/n)(a_n + a{n+1} + ... + a_{1n-1}) converges.

  1. #16
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    Re: Proving that (1/n)(a_n + a{n+1} + ... + a_{1n-1}) converges.

    Quote Originally Posted by vidomagru View Post
    Ok I see that and I understand that the coefficients follow \dfrac{2n+1}{n(n+1)} \le 1. Thus \sum_{k = n}^{2n-1} a_k \le na_n, \forall n. But where does this lead us now?
    The coefficients follow \dfrac{2n+1}{n(n+1)} when n is even. But not when it is odd. Anyway, if you can show that the coefficients are no greater than 1, then you can show that 0\le \sum_{n\ge 1} \dfrac{1}{n}\left(\sum_{k=n}^{2n-1} a_k \right) = \sum_{n\ge 1} b_na_n where b_n = \begin{cases}\dfrac{2n+1}{n(n+1)} & n\mbox{ is even} \\ ??? & n\mbox{ is odd}\end{cases} \le 1. Hence, \sum_{n\ge 1} b_na_n \le \sum_{n\ge 1} a_n
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  2. #17
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    Re: Proving that (1/n)(a_n + a{n+1} + ... + a_{1n-1}) converges.

    Quote Originally Posted by SlipEternal View Post
    The coefficients follow \dfrac{2n+1}{n(n+1)} when n is even. But not when it is odd. Anyway, if you can show that the coefficients are no greater than 1, then you can show that 0\le \sum_{n\ge 1} \dfrac{1}{n}\left(\sum_{k=n}^{2n-1} a_k \right) = \sum_{n\ge 1} b_na_n where b_n = \begin{cases}\dfrac{2n+1}{n(n+1)} & n\mbox{ is even} \\ ??? & n\mbox{ is odd}\end{cases} \le 1. Hence, \sum_{n\ge 1} b_na_n \le \sum_{n\ge 1} a_n
    I am not sure that I got that right. For example when n = 2, \dfrac{2(2) + 1}{2(2+1)} = \dfrac{5}{6} which is the coefficient for a_3.

    Also a_1 and a_2 appear once, a_3 and a_4 appear twice, a_5 and a_6 appear three times, a_7 and a_8 appear four times, etc. I'm not quite sure how to fit that into the coefficients since the number of a_n terms for any given n increases every two terms.
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  3. #18
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    Re: Proving that (1/n)(a_n + a{n+1} + ... + a_{1n-1}) converges.

    \begin{array}{lll}b_1 = 1, & b_2 = \dfrac{1}{2}, & b_3 = \dfrac{1}{2} + \dfrac{1}{3}, \\ \\ b_4 = \dfrac{1}{3} + \dfrac{1}{4}, & b_5 = \dfrac{1}{3} + \dfrac{1}{4} + \dfrac{1}{5}, & b_6 = \dfrac{1}{4} + \dfrac{1}{5} + \dfrac{1}{6} \\ \\ b_7 = \dfrac{1}{4} + \dfrac{1}{5} + \dfrac{1}{6} + \dfrac{1}{7}, & b_8 = \dfrac{1}{5} + \dfrac{1}{6} + \dfrac{1}{7} + \dfrac{1}{8}, & b_9 = \dfrac{1}{5} + \dfrac{1}{6} + \dfrac{1}{7} + \dfrac{1}{8} + \dfrac{1}{9} \end{array}

    So, let's try to find a recurrence relation for b_n:

    b_2 = b_1 - \dfrac{1}{1} + \dfrac{1}{2} = b_1 - \dfrac{1}{2}

    b_3 = b_2 + \dfrac{1}{3}

    b_4 = b_3 - \dfrac{1}{2} + \dfrac{1}{4} = b_3 - \dfrac{1}{4}

    b_5 = b_4 + \dfrac{1}{5}

    b_6 = b_5 - \dfrac{1}{3} + \dfrac{1}{6} = b_5 - \dfrac{1}{6}

    In general, this recursion seems to be b_{n+1} = b_n + \dfrac{(-1)^n}{n+1} with initial condition b_1 = 1.
    Thanks from vidomagru
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  4. #19
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    Re: Proving that (1/n)(a_n + a{n+1} + ... + a_{1n-1}) converges.

    So, if I am correct about b_n, then this is true:

    \sum_{n\ge 1} \dfrac{1}{n}(a_n + \ldots + a_{2n-1}) = \sum_{n\ge 1} \left(a_n\sum_{k = 1}^n \dfrac{(-1)^{k+1}}{k}\right)

    The sum \sum_{k = 1}^n \dfrac{(-1)^{k+1}}{k} is an alternating harmonic series, so it reaches a maximum value of 1 and a minimum value of \dfrac{1}{2}. Its limit as n\to \infty is \ln 2.

    So, again, if b_n is really a partial sum of the alternating harmonic series, then

    \dfrac{1}{2}\sum_{n\ge 1} a_n \le \sum_{n\ge 1} \dfrac{1}{n}(a_n + \ldots + a_{2n-1}) = \sum_{n \ge 1} \left(a_n\sum_{k = 1}^n \dfrac{(-1)^{k+1}}{k}\right) \le \sum_{n\ge 1} a_n

    So, it converges somewhere between 1/2 and 1 times the original series.
    Last edited by SlipEternal; October 15th 2013 at 10:55 PM.
    Thanks from vidomagru
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