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Math Help - A question about series

  1. #1
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    A question about series

    I've got this question I can't even think how to begin answering...

    Let a_1, a_2, ... be a monotonically non-increasing sequence of positive numbers, so that the series \sum_{n=1}^{\infty} a_n converges. Let S be the set of all numbers obtainable as \sum_{j=1}^{\infty} a_{n_j} for some series n_1 < n_2 < n_3 < ....
    Show that S is an interval iff for all n \in N,
    a_n \le \sum_{j=n+1}^{\infty} a_j .

    For which r > 0 does \sum r^n satisfy the condition?

    Thanks in advance...
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  2. #2
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    Quote Originally Posted by Aldarion View Post
    a_n \le \sum_{j=n+1}^{\infty} a_j .

    For which r > 0 does \sum r^n satisfy the condition?
    The last part of this question is not so bad...

    For what value of r does the following hold: r^n \leq \sum_{j=n+1}^{\infty} r^j ?

    Solution: Let r^n \leq \sum_{j=n+1}^{\infty} r^j =\frac{r^{n+1}}{1-r} . Solving, r \geq \frac12 . And as you know, \sum_{n=0}^{\infty} r^n converges iff r<1. So the answer to this part of the question is r\in[\frac12,1)

    For the first part of the question, consider this linear mapping... \phi: S\rightarrow[0,1] , \phi(s)=\phi(\sum_{i\in A}a_i)=\sum_{i\in A}\frac1{2^i} for some A\subset\mathbb{N}

    For example, \phi(a_1+a_3+a_5+a_7+...)=.10101010101_2=\frac23
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  3. #3
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    I'm sorry, but I still don't understand how does this linear mapping solve the first part of the question.

    Thanks for the second part though!
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  4. #4
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    Quote Originally Posted by Aldarion View Post
    Let a_1, a_2, ... be a monotonically non-increasing sequence of positive numbers, so that the series \sum_{n=1}^{\infty} a_n converges. Let S be the set of all numbers obtainable as \sum_{j=1}^{\infty} a_{n_j} for some series n_1 < n_2 < n_3 < ....
    Show that S is an interval iff for all n \in N,
    a_n \le \sum_{j=n+1}^{\infty} a_j .
    This is an "if and only if", so there are two parts.

    The "only if" part: suppose there exists n_0\in\mathbb{N}^* such that a_{n_0} > \sum_{j=n_0+1}^{\infty} a_j. Then the following numbers are in S : \alpha=\sum_{n=1}^{n_0-1} a_n + a_{n_0} and \beta=\sum_{n=1}^{n_0-1} a_n + \sum_{n=n_0+1}^\infty a_n, and \alpha>\beta. But the real numbers in (\beta,\alpha) do not belong to S (I let you prove this fact), so that S is not an interval.

    N.B. : I'm not sure whether \alpha is in S, it depends what is accepted as a subsequence. But it doesn't matter for this proof since s=\sum_{n=1}^\infty a_n is surely in S, and it is greater than \alpha. We have 0\in S and s\in S, and (\beta,\alpha)\subset[0,s] hence, to prove that S is not an interval, it suffices to show that some element of (\beta,\alpha) is not in S.


    The "if" part: suppose for all n, a_n \le \sum_{j=n+1}^{\infty} a_j. Obviously, since a_n\ge 0, the endpoints of S are 0 and s=\sum_{n=1}^\infty a_n. Let 0<x<s. We need to define a subsequence (a_{n_i})_i such that \sum_{i=1}^\infty a_{n_i}=x. Procede as follows: put a_1,\ldots,a_k in the subsequence, where k is the greatest index such that \sum_{i=1}^k a_i< x. Then wait for the first l>k such that \sum_{i=1}^k a_i + a_l<x, and put a_l,a_{l+1},\ldots,a_m in the subsequence, where m is the greatest index such that \sum_{i=1}^k a_i  + \sum_{i=l}^m a_i <x . And so on... We try to approach x from below using terms in the series. You have to show that this procedure makes sense, and that the subsequence you are making sums to x.

    That's the plan.
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