Prove that if an infinite series converges, then the associative property holds I'm self-studying from the book Understanding Analysis by Stephen Abbott and have no idea how to do exercise 2.5.2 on page 57.
The exercise is as follows:
Prove that if an infinite series converges, then the associative property holds. Assume $a_1+a_2 + a_3+a_4 + a_5+\cdots$ converges to a limit $L$ (i.e., the sequence of partial sums $(s_n) \to L$). [This sentence is already confusing me; I don't understand why if $(a_n) \to L$, this implies that $(s_n) \to L$?] Show that any regrouping of the terms
$$
(a_1 + a_2 + \cdots + a_{n_1}) + (a_{n_1+1} + \cdots + a_{n_2}) + (a_{n_2 + 1} + \cdots + a_{n_3}) + \cdots
$$
leads to a series that also converges to $L$.
Now, I'm aware that it is best to show what I've tried so far, but I have no idea how to get started. Any insight is much appreciated.
 A: by introducing the sequence $(n_k)_k$ which is an  extraction of the sequence $(n)_n$ the proof is easy: now we have $s_{n_k}=(a_1+\ldots+a_{n_1})+\ldots+(a_{n_1+1}+\ldots+a_{n_k})$, as $s_n\to L$ then $s_{n_k}\to L$ ($s_{n_k}$ is extraction of $(s_n)$ ). 
A: For your first confusion, $a_1 + a_2 + a_3+\cdots$ converges to a limit $L$ is just $\lim_{n\to \infty}s_n = L$ by definition. It does not mean $\lim_{n\to \infty}a_n = L$. 
Then to answer the question, $\lim_{n\to \infty}s_n = L$ means $\forall \epsilon >0$, there exists $N$ such that for all $n > N$, we have $|s_n - L |< \epsilon$.
Denote 
$b_1 = (a_1 + a_2 + \cdots + a_{n_1})$
$b_2 = (a_1 + a_2 + \cdots + a_{n_1}) + (a_{n_1+1} + \cdots + a_{n_2})$
$b_3 = (a_1 + a_2 + \cdots + a_{n_1}) + (a_{n_1+1} + \cdots + a_{n_2}) + (a_{n_2 + 1} + \cdots + a_{n_3}) $
Then you have that $b_k = s_{n_k}$ for some $n_k$ no less than $k$. So if $k>N$, $n_k \geq k >N$, then $|b_k-L| = |s_{n_k}-L| < \epsilon$. 
By definition, we have proven $b_k$ converges to $L$
