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Thread: Series convergence

  1. #1
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    Series convergence

    if a1+a2+a3.. decreasing sequence of positive numbers
    How to prove that if a1+a2+a3 ... converges then lim (if n to infinity) of na(subn)=0
    Last edited by drpawel; Nov 14th 2008 at 03:06 PM.
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  2. #2
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    \sum \frac{1}{n} is divergent, so for all \varepsilon>0, if n is enough big,
    |a_n|\leq \frac{\varepsilon}{n}, ie |n a_n|\leq \varepsilon.
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    I do not uderstand why we look at 1/n since it diverges,and in the problem series converges
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    Quote Originally Posted by drpawel View Post
    How to prove that if a1+a2+a3 ... converges then lim (if n to infinity) of na(subn)=0
    This result is not true without some further restriction on the terms a_n. For example, if a_n = \begin{cases}1/\sqrt n&\text{if $n=k^4$ for some integer $k$},\\0&\text{otherwise},\end{cases}
    Then \sum a_n = \sum_k 1/k^2, which converges. But if n=k^4 then na_n=\sqrt n, so a_n\not\to0.

    You need an extra condition such as requiring that the sequence (a_n) is decreasing, in order for the result to be true.
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    Thabk you for that correction.That sequence is indeed decreasing sequence of positive numbers.So,how to prove it.
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    Quote Originally Posted by drpawel View Post
    Yes, indeed the sequence is decreasing sequence of the positive numbers.How can i prove that now?
    I think you need to roll up your sleeves and wheel out some heavy-duty epsilons in order to prove this.

    Let \varepsilon>0. By the Cauchy criterion for convergence, there exists an integer N such that \sum_{j=m}^na_j<\varepsilon/2 whenever n>m>N. In particular, \sum_{j=N+1}^{N+p}a_j<\varepsilon/2 for all p>0. If the sequence (a_n) is decreasing, then pa_{N+p}\leqslant\sum_{j=N+1}^{N+p}a_j<\varepsilon/2, and therefore (N+p)a_{N+p}<\frac{N+p}p\varepsilon/2. But if p>N then \frac{N+p}p<2. It follows that if n>2N then na_n<\varepsilon. Therefore na_n\to0 as n\to\infty.
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    Quote Originally Posted by drpawel View Post
    and therefore
    If you have a specific follow-up question you will need to state it more clearly than this ....
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    Quote Originally Posted by mr fantastic View Post
    If you have a specific follow-up question you will need to state it more clearly than this ....
    I have some problems with my computer,I am trying to fix it, I did not do this by purpose.I appologise.
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    Quote Originally Posted by Opalg View Post
    I think you need to roll up your sleeves and wheel out some heavy-duty epsilons in order to prove this.

    Let \varepsilon>0. By the Cauchy criterion for convergence, there exists an integer N such that \sum_{j=m}^na_j<\varepsilon/2 whenever n>m>N. In particular, \sum_{j=N+1}^{N+p}a_j<\varepsilon/2 for all p>0. If the sequence (a_n) is decreasing, then pa_{N+p}\leqslant\sum_{j=N+1}^{N+p}a_j<\varepsilon/2, and therefore (N+p)a_{N+p}<\frac{N+p}p\varepsilon/2. But if p>N then \frac{N+p}p<2. It follows that if n>2N then na_n<\varepsilon. Therefore na_n\to0 as n\to\infty.
    Can you plese explain why (N+p)asub(N+p)<(N+p)/P, what about epsilon/2.Is it multiply time (N+p)/P.
    Plese explain a little bit since I am lost.
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  10. #10
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    Quote Originally Posted by drpawel View Post
    Can you plese explain why (N+p)asub(N+p)<(N+p)/P, what about epsilon/2.Is it multiply time (N+p)/P.
    Plese explain a little bit since I am lost.
    In the sum \sum_{j=N+1}^{N+p}a_j, each term is greater than or equal to the last term (since the sequence (a_j) is decreasing). So the sum is \geqslant  pa_{N+p}. That gives the inequality pa_{N+p}<\varepsilon/2.

    But we're interested in what happens to na_n. So it would be more useful to know something about (N+p)a_{N+p} rather than pa_{N+p}. That's the reason for multiplying both sides of the previous inequality by \frac{N+p}p, getting (N+p)a_{N+p}<\frac{N+p}p\varepsilon/2.

    I hope that makes it a bit clearer. As I said in my previous comment, this proof is fairly heavy-duty analysis. As far as I know, there are no easy shortcuts for proving this result.
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