Convergence of a sum of series I was reading the famous "Calculus" by Spivak and at the beginning of the chapter on infinite series, he states: "It's an easy exercise to prove that if both $\displaystyle \sum_{k=1}^{\infty}a_n$ and $\displaystyle \sum_{k=1}^{\infty}b_n$ exist, then $\displaystyle \sum_{k=1}^{\infty}a_n+b_n$=$\displaystyle \sum_{k=1}^{\infty}a_n$+$\displaystyle \sum_{k=1}^{\infty}b_n$". So I did it then. But after some pages he states that a rearrengament of the terms may change the limit. Then he proves that if $\displaystyle \sum_{k=1}^{\infty}|a_n|$ converges, any rearrengment of the terms wil converge to the same value. And so I started to think...How did I prove the asertion at the beginning of the chapter? It was just changing the order! Since $\displaystyle \sum_{k=1}^{\infty}a_n+b_n$ indeed means $\lim a_1+b_1+a_2+b_2+...+a_n+b_n$, it should(I thought) be the same than $\lim a_1+a_2+...+a_n+b_1+b_2+...+b_n=\displaystyle \sum_{k=1}^{\infty}a_n$+$\displaystyle \sum_{k=1}^{\infty}b_n$. So, where is the mistake? Should it be proved in another way?
 A: By definition, the series $\sum_{n=1}^{\infty}a_n$ exists and equals $A$ if and only if $$\lim_{n \to \infty}S_n=A$$where $S_n$ denotes the $n^{\text{th}}$ partial sum (i.e. $S_1=a_1$, $S_2=a_1+a_2$, etc.). Note that the $S_n$ form a sequence of real numbers. Now, assume that $\sum_{n=1}^{\infty}a_n=A$ and $\sum_{n=1}^{\infty}b_n=B$. We wish to show that $\sum_{n=1}^{\infty}(a_n+b_n)=A+B$. In other words, we wish to show that the limit of the partial sums of $\sum_{n=1}^{\infty}(a_n+b_n)$ is $A+B$. Let $S_{n}^A$ and $S_{n}^B$ denote that partial sums of $\sum_{n=1}^{\infty}a_n$ and $\sum_{n=1}^{\infty}b_n$ respectively. Fix $\epsilon>0$. By convergence, we know that there exists an $N\in \mathbb{N}$ such that $$n> N \Rightarrow |S_{n}^A-A|<\frac{\epsilon}{2}, |S_{n}^B-B|<\frac{\epsilon}{2}$$Now, note that the $n^{\text{th}}$ partial sum  of $\sum_{n=1}^{\infty}(a_n+b_n)$ is equivalent to $S_{n}^A+S_{n}^B$ (i.e. $S_n^{A+B}=S_n^A+S_n^B$, where $S_{n}^{A+B}$ denotes the $n^{\text{th}}$ partial sum of $\sum_{n=1}^{\infty}(a_n+b_n)$). 

Now, using the same $N$ as above we have that for $n>N$, $$|S_n^{A+B}-(A+B)|=|(S_n^A-A)+(S_n^B-B)|\leq |S_n^A-A|+|S_n^B-B|<\frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon$$Thus, by the definition of convergence, the sequence $S_{n}^{A+B}$ of partial sums of $\sum_{n=1}^{\infty}(a_n+b_n)$  converges to $A+B$ and therefore $\sum_{n=1}^{\infty}(a_n+b_n)=A+B$
A: In other words, you did not make any mistake. We know for all $K$: 
$$ \underbrace{\sum_{n=1}^K [a_n + b_n]}_{r_K} = \underbrace{\sum_{n=1}^K a_n}_{w_K} + \underbrace{\sum_{n=1}^K b_n}_{z_K} $$
So $r_K = w_K + z_K$ for all $K$. 
You do not know yet if $r_K$ has a limit as $K\rightarrow\infty$, but if you are told that $w_K$ and $z_K$ both have a finite limit, we know by basic limit facts that 
$$\lim_{K\rightarrow\infty} r_K = \lim_{K\rightarrow\infty}w_K + \lim_{K\rightarrow\infty} z_K$$
The reason this does not have any divergence problems, even if $a_n = (-1)^n/n$, is that you are still taking the values $\{w_1, w_2, w_3, ...\}$ and $\{z_1, z_2, z_3, ...\}$ in the same relative order, both when they are grouped together and when they are grouped separately. 

Another way of interpreting this that may help:  Suppose $\{a_n\}_{n=1}^{\infty}$ is a real valued sequence such that $\sum_{n=1}^{\infty}|a_n|=\infty$. This does not necessarily mean that every different ordering for summing the $a_n$  terms will lead to a different sum.  Many different orders will in fact lead to the same sum. In particular, suppose  you fix an order $\{a_1, a_2, a_3, ...\}$.  You now separate the terms into two groups without changing the relative ordering within each group: Suppose the first group pulls out a subsequence of the $a_n$ terms that happen to sum to a finite number $x$ (in that relative order). Suppose the remaining terms (which form the second group) also happens to sum to a finite number $y$ (in that relative order).  Then  the original sum will converge to $x+y$ in any order that alternates between the first and second groups, provided that you are not changing the relative ordering within each group. 
For example, you can sum the first 4 terms of group 1, then the first 8 terms of group 2, then the next 3 terms of group 1, then the next 100 terms of group 2, and so on. It will give you the same result $x+y$, provided your way of alternating eventually includes all terms and preserves the relative ordering of terms for each group. 
