# (Real Analysis) If a series converges absolutely, then any rearrangement of this series converges to the same limit. Why is the following proof valid?

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?

• 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 $\sum_{k = 1}^{\infty} a_k = A$? – Martin R May 30 at 18:36
• @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? – pup2089 May 30 at 18:48

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.
• 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? – pup2089 May 30 at 18:44
• 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 $\displaystyle\sum_{k=1}^\infty a^k=s$. – José Carlos Santos May 30 at 18:58
• 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? – pup2089 May 30 at 19:18
• 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 $B$. – José Carlos Santos May 30 at 19:42