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Show that a conditionally convergent series has a rearrangement converging to $+\infty$


Thoughts:

  1. A conditionally convergent series is a series that converges but not absolutely converges $\lim_{m\to\infty} \sum_{n=0}^{m} a_n$ exists, $\sum_{n=0}^{\infty} |a_n|=\infty$

  2. if $\sum_{n=0}^{\infty} a_n$ is a conditionally convergent series, then for every real number $L$, there is a rearrangement that converges to $L$

Since we are given $\sum_{n=0}^{\infty} a_n$ converges conditionally. Intuitively, we obtain one positive and one negative series, which should be converge to the same limit.

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    $\begingroup$ The last statement is somewhat vague.. The positive and negative series diverge. Unless you mean to say one converge to $\infty$ and the other to $-\infty$. $\endgroup$
    – Maesumi
    Feb 27, 2013 at 23:28
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    $\begingroup$ Do you know how to make a conditionally convergent series sum to a given $L\in\mathbb{R}$? $\endgroup$
    – Stahl
    Feb 27, 2013 at 23:29
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    $\begingroup$ Suppose your series, when separated into positive and negative parts is formally $\sum p_i$+$\sum n_j$. Add enough $p_i$'s until you exceed $1$ then add just $n_1$. Then add enough $p_i$'s until you exceed $2$ and add $n_2$ and so on add enough positive ones until you exceed $k$ and then add $n_k$. The resulting series diverges to $+\infty$ and has all original terms in it. $\endgroup$
    – Maesumi
    Feb 27, 2013 at 23:34
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    $\begingroup$ I suppose @Paul wants a formal proof. $\endgroup$
    – Maesumi
    Feb 27, 2013 at 23:40
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    $\begingroup$ Suppose $L\in\mathbb{R}^+$. The way we normally construct the rearrangement is by arranging the positive terms $p_i$ and negative terms $n_i$ so that $\left|p_i\right|\geq\left|p_{i+1}\right|$, and $\left|n_i\right|\geq\left|n_{i+1}\right|$. Then we add $p_1 + p_2 + \ldots + p_n$ such that $n$ is the smallest integer with $p_1 + p_2 + \ldots + p_n \geq L$. Then add $n_1$. Proceed to add more $p_i$'s until the sum again exceeds $L$, and repeat. This will give you a series that sums to $L$. @Maesumi's argument is an adaption to the case $L = \infty$. $\endgroup$
    – Stahl
    Feb 27, 2013 at 23:47

2 Answers 2

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This is a particular case of the astonishing Riemann Series Theorem . Note that you can rearrange a conditional convergent series in such a way as to make the rearranged series converge to whatever you want.

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  • $\begingroup$ would you explain a little bit more? $\endgroup$
    – Paul
    Feb 28, 2013 at 1:40
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    $\begingroup$ Read the theorem in the link: if $\,\sum a_n\,$ is a conditionally convergent series, then for any $\,\alpha\in\Bbb R\cup\{\pm\infty\}\,$ there exists a permutation $\,\sigma\in S_{\Bbb N}\,$ s.t. $\,\sum a_{\sigma{n}}=\alpha\,$ ...This means that upon being given any real number or $\,\pm\infty\,$ , you can arrange the series's terms as to get a new series that converges to that...amazing, uh?! $\endgroup$
    – DonAntonio
    Feb 28, 2013 at 2:20
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We have that $\sum_n \frac{(-1)^n}{n}-\frac{(-1)^n}{n} = 0$, but $\sum_n \frac{2}{n} = \infty$. Then $(c_n)$ where

\begin{align} c_0 &= 1, \\ c_1 &= -1, \\ f(n) &= \max_{k < n} \left\{\frac{1}{c_k}\right\}+1, \\ g(n) &= \min_{k < n} \left\{\frac{1}{c_k}\right\}-1, \\ c_n &= \begin{cases} \frac{1}{f(n)} & \text{if } \sum_{k<n} c_k < -g(n)\\ \frac{1}{g(n)} & \text{if } \sum_{k<n} c_k \geq -g(n) \end{cases} \end{align}

would be your desired sequence. Have fun ;-)

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