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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.

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    $\begingroup$ When he says that the series $a_1+a_2+\cdots$ converges to L, this means that $(s_n)\rightarrow L$, not that $(a_n)\rightarrow L$. To show the exercise, see what you can say about the partial sums for the series you get after grouping the terms. $\endgroup$ – user84413 Aug 15 '14 at 20:12
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    $\begingroup$ The $n$th term in the sequence of partial sums is $$s_n = \sum_1^n a_n.$$ Does that help? $\endgroup$ – John Aug 15 '14 at 20:17
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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 $L$

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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 $(s_n)$ ).

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  • $\begingroup$ But how do you know that all subsequences of a sequence converge to the same limit as the sequence? $\endgroup$ – Elliot Gorokhovsky Aug 16 '14 at 0:02
  • $\begingroup$ @RenéG I believe this is a well-known theorem. This is proven in theorem 2.5.2 in the book I've mentioned in my question: Subsequences of a convergent sequence converge to the same limit as the original sequence. $\endgroup$ – Hunter Aug 16 '14 at 0:20

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