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

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

    Let \sum\limits_{n=1}^\infty a_n be a convergent series in which a_n \ge 0, \forall n=1,2, .... Prove that the series \sum\limits_{n=1}^\infty \frac{1}{n} (a_n + a_{n+1} + ... + a_{2n-1}) converges as well.


    How should I start?
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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
    Let \sum\limits_{n=1}^\infty a_n be a convergent series in which a_n \ge 0, \forall n=1,2, .... Prove that the series \sum\limits_{n=1}^\infty \frac{1}{n} (a_n + a_{n+1} + ... + a_{2n-1}) converges as well.


    How should I start?
    Proving that (1/n)(a_n + a{n+1} + ... + a_{1n-1}) converges.-untitled2.gif
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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
    Let \sum\limits_{n=1}^\infty a_n be a convergent series in which a_n \ge 0, \forall n=1,2, .... Prove that the series \sum\limits_{n=1}^\infty \frac{1}{n} (a_n + a_{n+1} + ... + a_{2n-1}) converges as well.


    How should I start?
    Maybe this might simplify something:
    \sum\limits_{n=1}^\infty \frac{1}{n} (a_n + a_{n+1} + ... + a_{2n-1}) = \sum\limits_{n=1}^\infty \frac{1}{n} \left(\sum_{k = 1}^{2n-1} a_k - \sum_{k = 1}^{n-1} a_k \right)
    Last edited by SlipEternal; October 13th 2013 at 08:01 PM.
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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
    Maybe this might simplify something:
    \sum\limits_{n=1}^\infty \frac{1}{n} (a_n + a_{n+1} + ... + a_{2n-1}) = \sum\limits_{n=1}^\infty \frac{1}{n} \left(\sum_{k = 1}^{2n-1} a_k - \sum_{k = 1}^{n-1} a_k \right)
    So I understand how you get there I think. To parse this:

    I know that \sum\limits_{n=1}^\infty \frac{1}{n} diverges.

    I believe that (\sum_{k = 1}^{2n-1} a_k and \sum_{k = 1}^{n-1} a_k) converge since \sum\limits_{n=1}^\infty a_n in convergent for all n = 1,2,.... Is that right?

    But I do not know where to go from there?
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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
    So I understand how you get there I think. To parse this:

    I know that \sum\limits_{n=1}^\infty \frac{1}{n} diverges.

    I believe that (\sum_{k = 1}^{2n-1} a_k and \sum_{k = 1}^{n-1} a_k) converge since \sum\limits_{n=1}^\infty a_n in convergent for all n = 1,2,.... Is that right?

    But I do not know where to go from there?
    Let S_k = \sum_{n = 1}^k a_n. Since the series converges absolutely, the sequence S_k converges (and is Cauchy)[/tex]. Rewriting the series you are given:

    \sum_{n = 1}^\infty \dfrac{S_{2n-1} - S_{n-1}}{n}

    Writing out a few elements of this series, you have S_1 + \dfrac{S_3 - S_1}{2} + \dfrac{S_5 - S_2}{3} + \dfrac{S_7 - S_3}{4} + \ldots

    Collecting like terms, we have \dfrac{S_1}{2} - \dfrac{S_2}{3} + \dfrac{S_3}{4} - \dfrac{S_4}{5} \pm \ldots

    So, the general form of the series seems to be \sum_{n \ge 1} (-1)^{n+1} \dfrac{S_n}{n+1}. Use induction to prove that this is true, and then use the Alternating Series Test to prove that it converges conditionally.
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    Re: Proving that (1/n)(a_n + a{n+1} + ... + a_{1n-1}) converges.

    Ok I follow your logic up until here:

    Quote Originally Posted by SlipEternal View Post
    So, the general form of the series seems to be \sum_{n \ge 1} (-1)^{n+1} \dfrac{S_n}{n+1}. Use induction to prove that this is true, and then use the Alternating Series Test to prove that it converges conditionally.
    If I try to prove this is true using induction I have:

    \sum_{n \ge 1} (-1)^{n+1} \dfrac{S_n}{n+1} evaluated at n=1 gives (-1)^{1+1} \dfrac{S_1}{1+1} = \dfrac{a_1}{2}.

    However this is not equal to \sum\limits_{n=1}^\infty \dfrac{1}{n} (a_n + a_{n+1} + ... + a_{2n-1}) evaluated at n=1, i.e., (a_1 + a_2 + ...  + a_1).
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    Re: Proving that (1/n)(a_n + a{n+1} + ... + a_{1n-1}) converges.

    Hmm, that's a good point. Maybe you can show that for any \varepsilon>0, there exists N such that for all n\ge N, the partial sum of the first n terms are no more than \varepsilon apart?
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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
    Hmm, that's a good point. Maybe you can show that for any \varepsilon>0, there exists N such that for all n\ge N, the partial sum of the first n terms are no more than \varepsilon apart?
    I am not quite sure how to apply that, but I thought of something else instead that might be helpful. Isn't (a_n + a_{n+1} + ... + a_{2n-1}) a subsequence of \sum\limits_{n=1}^\infty a_n which is convergent so any subsequence would also be convergent?

    Or because (a_n + a_{n+1} + ... + a_{2n-1}) is a partial sum of a convergent series \sum\limits_{n=1}^\infty a_n, (a_n + a_{n+1} + ... + a_{2n-1}) also converges.

    Then we would just need to apply \sum\limits_{n=1}^\infty \dfrac{1}{n}. Not sure how to do that?
    Last edited by vidomagru; October 15th 2013 at 06:37 PM.
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    Re: Proving that (1/n)(a_n + a{n+1} + ... + a_{1n-1}) converges.

    A subsequence is very different from a partial sum. A sum of partial sums may not converge. But, I think I figured it out.

    \sum_{n\ge 1} \dfrac{1}{n}\left( \sum_{k = n}^{2n-1}a_k \right)

    Now, can you show that \sum_{k = n}^{2n-1} a_k \le na_n for almost all n?

    Since \sum_{n\ge 1} a_n converges and 0\le a_n for all n, it must be that \lim_{n\to \infty} a_n = 0. So what I am getting at, \underbrace{a_n+ a_{n+1} + \ldots + a_{2n-1}}_{n\mbox{ terms}} \le \underbrace{a_n + a_n + \ldots + a_n}_{n\mbox{ terms}}
    Last edited by SlipEternal; October 15th 2013 at 07:14 PM.
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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
    A subsequence is very different from a partial sum. A sum of partial sums may not converge. But, I think I figured it out.

    \sum_{n\ge 1} \dfrac{1}{n}\left( \sum_{k = n}^{2n-1}a_k \right)

    Now, can you show that \sum_{k = n}^{2n-1} a_k \le na_n for almost all n?
    s
    Last edited by vidomagru; October 15th 2013 at 07:26 PM.
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    Re: Proving that (1/n)(a_n + a{n+1} + ... + a_{1n-1}) converges.

    Also, writing out the terms of the series, we have:
    a_1 + \dfrac{a_2}{2} + \dfrac{5a_3}{6} + \dfrac{7a_4}{12} + \dfrac{47a_5}{60} + \ldots

    Can you show that the coefficient of a_n is always no greater than 1?
    Last edited by SlipEternal; October 15th 2013 at 07:24 PM.
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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
    A subsequence is very different from a partial sum. A sum of partial sums may not converge. But, I think I figured it out.

    \sum_{n\ge 1} \dfrac{1}{n}\left( \sum_{k = n}^{2n-1}a_k \right)

    Now, can you show that \sum_{k = n}^{2n-1} a_k \le na_n for almost all n?

    Since \sum_{n\ge 1} a_n converges and 0\le a_n for all n, it must be that \lim_{n\to \infty} a_n = 0. So what I am getting at, \underbrace{a_n+ a_{n+1} + \ldots + a_{2n-1}}_{n\mbox{ terms}} \le \underbrace{a_n + a_n + \ldots + a_n}_{n\mbox{ terms}}
    I don't quite understand how \underbrace{a_n+ a_{n+1} + \ldots + a_{2n-1}}_{n\mbox{ terms}} is n terms? Also I cannot get induction to work for

    Quote Originally Posted by SlipEternal View Post
    Now, can you show that \sum_{k = n}^{2n-1} a_k \le na_n for almost all n?
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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
    Also, writing out the terms of the series, we have:
    a_1 + \dfrac{a_2}{2} + \dfrac{5a_3}{6} + \dfrac{7a_4}{12} + \dfrac{47a_5}{60} + \ldots

    Can you show that the coefficient of a_n is always no greater than 1?
    I also don't think I quite understand how you expanded the series.
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    Re: Proving that (1/n)(a_n + a{n+1} + ... + a_{1n-1}) converges.

    \sum_{n\ge 1} \dfrac{1}{n}\left(\sum_{k = n}^{2n-1}a_k\right)

    \dfrac{a_1}{1} + \dfrac{a_2 + a_3}{2} + \dfrac{a_3 + a_4 + a_5}{3} + \dfrac{a_4 + a_5 + a_6 + a_7}{4} + \dfrac{a_5 + a_6 + a_7 + a_8 + a_9}{5} + \ldots

    a_1 shows up in the first term only. a_2 shows up in the second term only. a_3 shows up in terms 2 and 3. a_4 shows up in terms 3 and 4. a_5 shows up in terms 3, 4, and 5. a_6 shows up in terms 4, 5, and 6. Seeing a pattern?

    Also, this expansion should show you how each term is a sum of n elements over n.
    Last edited by SlipEternal; October 15th 2013 at 07:51 PM.
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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
    \sum_{n\ge 1} \dfrac{1}{n}\left(\sum_{k = n}^{2n-1}a_k\right)

    \dfrac{a_1}{1} + \dfrac{a_2 + a_3}{2} + \dfrac{a_3 + a_4 + a_5}{3} + \dfrac{a_4 + a_5 + a_6 + a_7}{4} + \dfrac{a_5 + a_6 + a_7 + a_8 + a_9}{5} + \ldots

    a_1 shows up in the first term only. a_2 shows up in the second term only. a_3 shows up in terms 2 and 3. a_4 shows up in terms 3 and 4. a_5 shows up in terms 3, 4, and 5. a_6 shows up in terms 4, 5, and 6. Seeing a pattern?
    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?
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