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The following proof is from Stephen Abbott's Understanding Analysis 2nd edition, page 75. I have put in bold the assumption which I don't understand.

Proof: Assume $\sum\limits_{k = 1}^{\infty} a_k$ converges absolutely to $A$, and let $\sum\limits_{k = 1}^{\infty} b_k$ be a rearrangement of $\sum\limits_{k = 1}^{\infty} a_k$. Let's use $s_n$ to denote the partial sums of the original series and $t_m$ for the partial sums of the rearranged series. Thus, we want to show that $(t_m) \to A$.

Let $\epsilon > 0$. By hypothesis, $\boldsymbol{(s_n) \to A}$, so choose $N_1$ such that $|s_n - A| < \frac{\epsilon}{2}$ for all $n \geq N_1$. Because the convergence is absolute, we can choose $N_2$ so that $\sum\limits_{k = m + 1}^{n} |a_k| < \frac{\epsilon}{2}$ for all $n > m \geq N_2$. Now take $N = \max \{N_1, N_2\}$. We know that the finite set of terms $\{a_1, \ldots, a_N\}$ must all appear in the rearranged series, and we want to move far enough out in the series $\sum\limits_{n = 1}^{\infty} b_n$ so that we have included all of these terms. Thus, choose $M = \max \{f(k) : 1 \leq k \leq N \}$.

It should now be evident that if $m \geq M$, then $(t_m - s_N)$ consists of a finite set of terms, the absolute values of which appear in the tail $\sum\limits_{k = N + 1}^{\infty} |a_k|$. Our choise of $N_2$ earlier then guarantees $|t_m - s_N| < \frac{\epsilon}{2}$, and so $|t_m - A| = |t_m - s_N + s_N - A| \leq |t_m - s_N| + |s_N - A| < \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon$ whenever $m \geq M$.

My Question: Why does it say "By hypothesis, $(s_n) \to A$"? I thought we are assuming the series converges absolutely to A. In other words, we are assuming that $\sum\limits_{k = 1}^{\infty} |a_k| = A$, not that $\sum\limits_{k = 1}^{\infty} a_k = A$, right?

Previously, in the book, we only proved that if a series converges absolutely, then the original series converges as well (Absolute Convergence Test). We didn't show that if a series converges absolutely to some specific limit A, then the original series also converges to A. Is it actually true that if a series converges absolutely to A, then the original series also converges to A?

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    $\begingroup$ Isn't it apparent from the text that “$\sum_{k = 1}^{\infty} a_k$ converges absolutely to $A$” means that $\sum_{k = 1}^{\infty} a_k$ converges absolutely, and $\sum_{k = 1}^{\infty} a_k = A$? $\endgroup$ – Martin R May 30 at 18:36
  • $\begingroup$ @MartinR The definition of an absolutely convergent series in the book just says "If $\sum\limits_{k = 1}^{\infty} |a_k|$ converges, then we say that the original series $\sum\limits_{k = 1}^{\infty} a_k$ converges absolutely." Where do we get $\sum\limits_{k = 1}^{\infty} a_k = A$ from this definition? $\endgroup$ – pup2089 May 30 at 18:48
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You are misunderstanding the meaning of “the series converges absolutely to $A$”. It means two things:

  1. the series $\displaystyle\sum_{k=1}^\infty a_k$ converges and its sum is $A$;
  2. the series $\displaystyle\sum_{k=1}^\infty|a_k|$ converges.
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  • $\begingroup$ The definition of an absolutely convergent series in the book just says "If $\sum\limits_{k = 1}^{\infty} |a_k|$ converges, then we say that the original series $\sum\limits_{k = 1}^{\infty} a_k$ converges absolutely." Do you know of another definition somewhere online I can see? $\endgroup$ – pup2089 May 30 at 18:44
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    $\begingroup$ That is the usual definition of absolutely convergent series. But note that when we say that a series $\displaystyle\sum_{k=1}^\infty a^k$ converges to $s$ what we mean is that $\displaystyle\sum_{k=1}^\infty a^k=s$. $\endgroup$ – José Carlos Santos May 30 at 18:58
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    $\begingroup$ Also, take a look here. $\endgroup$ – José Carlos Santos May 30 at 18:59
  • $\begingroup$ Okay, thank you. What if we had the case where $\sum\limits_{k = 1}^{\infty} |a_k| = A$ but $\sum\limits_{k = 1}^{\infty} a_k = B$? Would we then say $\sum\limits_{k = 1}^{\infty} a_k$ converges absolutely to B, in this case? $\endgroup$ – pup2089 May 30 at 19:18
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    $\begingroup$ Yes, if $\displaystyle\sum_{k=1}^\infty|a_k|=A$ and $\displaystyle\sum_{k=1}^\infty a_k=B$, we say that $\displaystyle\sum_{k=1}^\infty a_k$ converges absolutely to $B$. $\endgroup$ – José Carlos Santos May 30 at 19:42

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