Conditionally converges and rearrangement

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.

• 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 $-\infty$. Feb 27 '13 at 23:28
• Do you know how to make a conditionally convergent series sum to a given $L\in\mathbb{R}$? Feb 27 '13 at 23:29
• 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. Feb 27 '13 at 23:34
• I suppose @Paul wants a formal proof. Feb 27 '13 at 23:40
• 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 $L = \infty$. Feb 27 '13 at 23:47

• 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?! Feb 28 '13 at 2:20
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