Prove that absolute convergence implies unconditional convergence In the proof of "absolute convergence implies unconditional convergence" for a convergent series $\sum_{n=1}^{\infty}a_n$, we take a partial sum of first $n$ terms of both the original series ($S_n$) and rearranged series ($S_n'$) and compare them. Because the original series converges, we get some $N$ from Cauchy-criteria. Now if I choose "$n$ large enough" such that $\{a_1,a_2,\dots,a_{N-1}\} \subseteq \{a_1',a_2',\dots,a_{n}'\}$. Then if we compare both the partial sums. the remaining $a_i$s are all for $i \geq N$, but still some $a_i'$s are remaining. The book claims that $$|\sum_{i=N}^{n}a_i-\sum_{i=N}^{n}a_i'| \leq |\sum_{i=N}^{n}|a_i||.$$ I could not not understand how these $a_i'$s are getting removed.
An infinite series is called "unconditionally convergent" if every rearrangement of it converges to the same value. If the sequence is absolutely convergent, then it can be shown that all converges to the same value, in fact, the above theorem proves that.
 A: If your text actually writes

$|\sum_{i=N}^{n}a_i-\sum_{i=N}^{n}a_i'| \leq |\sum_{i=N}^{n}|a_i||$

then it is indeed mistaken.  (Note also that even on its own the right hand side is strangely written: why do we need the outside absolute value?)  For instance, suppose $a_N = \ldots = a_n = 0$.  Then the inequality implies $\sum_{i=N}^n a_i' = 0$, but the assumptions do not give us that.
I would suggest a somewhat different proof, namely the one which is given in Theorem 14.7 of these notes.
Put $A = \sum_{n=0}^{\infty} a_n$.  Fix $\epsilon > 0$ and let $N_0 \in \mathbb{N}$ be such that  $\sum_{n=N_0}^{\infty} |a_n| < \epsilon$.  Let $M_0 \in \mathbb{N}$ be
sufficiently large so that the terms $a_{\sigma(0)},\ldots,a_{\sigma(M_0)}$
include all the terms $a_0,\ldots,a_{N_0-1}$ (and possibly others).  Then
for all $M \geq M_0$, $$|\sum_{n=0}^{M} a_{\sigma(n)} - A| = |\sum_{n=0}^{M} a_{\sigma(n)} - 
\sum_{n=0}^{\infty} a_n| \leq \sum_{n=N_0}^{\infty} |a_n| < \epsilon.$$
Indeed: by our choice of $M$ we know that the terms
$a_0,\ldots,a_{N_0-1}$ appear in both $\sum_{n=0}^M a_{\sigma(n)}$ and
$\sum_{n=0}^{\infty} a_n$ and thus get cancelled;
some further terms may or may not be cancelled, but by applying the triangle inequality and summing the absolute values we get an upper bound by assuming no further cancellation.  This shows $\sum_{n=0}^{\infty} a_{\sigma(n)} = \lim_{M \rightarrow \infty} \sum_{n=0}^M a_{\sigma(n)} = A$.
