As part of self-study, I am working on exercise 2.5.3 (a) of Understanding Analysis by Stephen Abbott (2015).
The proof looks to me to be a simple application of the fact that subsequences of a convergent sequence tend to the same limit as said sequence. My primary difficulty here is how to write up a formal proof the exercise requests in a way that is explicit and doesn't feel wanting.
Context.
Exercise 2.5.3.
(a) Prove that if an infinite series converges, then the associative property holds. Assume $a_1 + a_2 + a_3 + a_4 + a_5 + \dots \,$ converges to a limit $L$ (i.e., the sequence of partial sums $(s_m) \rightarrow L$). Show that any regrouping of the terms
$$(a_1 + a_2 + \dots + a_{n_1}) + (a_{n_1 + 1} + \dots + a_{n_2}) + (a_{n_2 + 1} + \dots a_{n_3}) + \dots $$leads to a series that converges to $L$.
Attempted proof.
Assume that $\sum^{\infty}_{n=1} a_n = L$. So the sequence of partial sums $(s_m) \rightarrow L$. We use theorem 2.5.2 that subsequences of a convergent sequence converge to the same limit as the original sequence. Therefore the subsequence of partial sums $(s_{n_k})$, where $k \in \mathbb{N}$ and the $n_k$ are increasing natural numbers, will also converge to the same limit $L$. Hence
$$(a_1 + a_2 + \dots + a_{n_1}) + (a_{n_1 + 1} + \dots + a_{n_2}) + (a_{n_2 + 1} + \dots a_{n_3}) + \dots = L.$$
Question.
Please bear in mind that I am new to proof writing and analysis. Therefore I need to be more explicit rather than less explicit in my proof write-ups. To me the above attempted proof feels wanting, in that it seems to "lack grip" on the crux of what is being asked, which is about associativity.
Comparing this to the solution posted in the thread below by Petite Etincelle, I understand I could be more explicit in including references to the standard definitions of convergence. But where I am confused is the introduction of an additional sequence $(b_k)$, which makes the answer more explicit, but seems redundant in that it is no different to $(s_{n_k})$
Is the inclusion of the sequence $(b_k)$ redundant or necessary at my level?
Related solution.
The solution to the same exercise in this thread, is:
Then to answer the question, $\lim_{n \rightarrow \infty} s_n = L$ means $\forall \epsilon > 0$, there exists $N$ such that for all $n>N$, we have $\lvert s_n – L \lvert < \epsilon$.
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 $\vert b_k – L \vert = \vert s_{n_k} − L \vert < \epsilon$.
By definition, we have proven $b_k$ converges to $L$.
Best Answer
The problem is that a subsequence $(s_{n_k})$ of the sequence of partial sums $(s_m)$ doesn't reflect the grouping of the terms.
Rather than saying the sequence of (one-term spaced) partial sums $a_1,~a_1+a_2,~a_1+a_2+a_3,\ldots$ converges, you're saying that the sequence of ($n_k$-term spaced) partial sums $$a_1+\ldots+a_{n_1}\\a_1+\ldots_+a_{n_1}+a_{n_1+1}+\ldots+a_{n_2}\\a_1+\ldots+a_{n_1}+a_{n_1+1}+\ldots+a_{n_2}+a_{n_2+1}+\ldots+a_{n_3}\\\vdots$$
converges, which as you can see, doesn't reflect on the grouping of terms at all.
What you need to do is to show that the sequence
$$(a_1+\ldots+a_{n_1})\\(a_1+\ldots_+a_{n_1})+(a_{n_1+1}+\ldots+a_{n_2})\\(a_1+\ldots+a_{n_1})+(a_{n_1+1}+\ldots+a_{n_2})+(a_{n_2+1}+\ldots+a_{n_3})\\\vdots$$
converges, for which you first need to enumerate the groups first, ie, write it as
$$s_{n_1}\\s_{n_1}+(s_{n_2}-s_{n_1})\\ s_{n_1}+(s_{n_2}-s_{n_1})+(s_{n_3}-s_{n_2})\\\vdots$$
from which the construction of $b_k$ becomes apparent, and why it is necessary for a rigorous (and constructive) proof of the claim, at any level.