A question on the convergence of series I was reading Protter's "Probability Essentials" and the author says in the fourth chapter that the convergence of series depends on the order in which they are enumerated. I cannot think of a situation in which by changing the enumeration , a converging series might cease to converge or vice-versa. Could someone please give me an example?
Thank you
 A: $$1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...=\sum_{n=1}^\infty (-1)^{n-1} \frac{1}{n}$$
Pick a real number $L$. 
Rearange the series the following way: pick the first positive terms until the sum exceeds $L$. Pick now the first negative terms so that, added to the first sum you get back under $L$.
Repeat: Whenever when above $L$, pick the first left negative terms until the sum is under $L$.  Whenever when below $L$, pick the first left positive terms until the sum is over $L$. 
This is possible because 
$$1+\frac{1}{3}+\frac{1}5+...=\frac{1}{2}+\frac{1}{4}+...=\infty
$$
This can always be done with series which are convergent BUT not absolutely convergent.
A: It could happen when you deal with series which are not absolutely convergent.
Take for example
$$
\sum_{n=1}^{+\infty}\frac{(-1)^{n+1}}n
$$
we know that this series converges by alternating test. Call $S$ the sum.
Look at the first terms:
$$
1-\frac12+\frac13-\frac14+\frac15-\frac16+\frac17-\frac18+\dots
$$
and rearrange them as follows:
$$
\left(1+\frac13-\frac12\right)+\left(\frac15+\frac17-\frac14\right)+\left(\frac19+\frac1{11}-\frac16\right)
$$
whose partial sums are, for short
$$
\widetilde S_{3n}=\sum_{k=1}^{n}\underbrace{\left({\frac1{4k-3}+\frac1{4k-1}-\frac1{2k}}\right)}_{=:a_k}
$$
now you can check easily that $a_k>0$, hence $\widetilde S_{3n}$ has limits $\widetilde S\in]0+\infty]$, but you can check as well that $a_k\sim\frac1{k^2}$ so the series converges, i.e. $\widetilde S\neq+\infty$.
Let's finally prove that $S\neq\widetilde S$. Observe that you can write $S$ in the two following ways:
$$
S=\sum_{k=1}^{+\infty}\frac1{2k-1}-\frac1{2k}=:A
$$
and, similarly
$$
S=\sum_{k=1}^{+\infty}\left(\frac1{4k-3}-\frac1{4k-2}+\frac1{4k-1}-\frac1{4k}\right)=:B
$$
thus
$$
\frac32S=\frac12A+B
$$
and this last one, with a simple computation reveals to be $\widetilde S$.
Hence $\frac32S=\widetilde S$, i.e. $S\neq\widetilde S$ as wanted.
This is an example of a phenomenon explained by Riemann in the following
Theorem: let $\sum_na_n$ a convergent but not absolutely convergent real series. Then, taking $\alpha,\beta\in\Bbb R\cup\{\pm\infty\}$, there exists a bijection $\sigma:\Bbb N\to\Bbb N$ such that, having called
$$
S_n:=\sum_{k=0}^{n}a_{\sigma(k)}
$$
we get
$$
\liminf_nS_n=\alpha\;\;\;\;\;\;\limsup_nS_n=\beta
$$
i.e. we can reorder the term of the series to get whatever sum we want!
