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

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

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

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

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

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

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

    For example, $\displaystyle \phi(a_1+a_3+a_5+a_7+...)=.10101010101_2=\frac23$
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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 $\displaystyle a_1, a_2, ... $ be a monotonically non-increasing sequence of positive numbers, so that the series $\displaystyle \sum_{n=1}^{\infty} a_n$ converges. Let S be the set of all numbers obtainable as $\displaystyle \sum_{j=1}^{\infty} a_{n_j}$ for some series $\displaystyle n_1 < n_2 < n_3 < ...$.
    Show that S is an interval iff for all $\displaystyle n \in N$,
    $\displaystyle a_n \le \sum_{j=n+1}^{\infty} a_j$ .
    This is an "if and only if", so there are two parts.

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

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


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

    That's the plan.
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