Justification.
Let
\(\nseq{s}\) and
\(\nseq{t}\) represent the sequences of partial sums of
\(\sum a_k\) and
\(\sum a_{\varphi(k)}\text{,}\) respectively, and let
\(S\) represent the value of
\(\sum a_k\text{,}\) so that
\(\nseq{s} \to S\text{.}\) We would like to verify that
\(\nseq{t} \to S\) as well. Using
Fact 22.3.31, we will instead verify that
\(\seq{t_n - S} \to 0\text{.}\) Following
Corollary 22.3.32, consider an arbitrary half-open range
\([0,c)\text{.}\) Our goal is to verify that some tail of the absolute sequence
\(\seq{\abs{t_n - S}}\) is completely contained in this range.
As \(\nseq{s} \to S\text{,}\) then also \(\seq{s_n - S} \to 0\text{,}\) and there exists some tail \(\absseqtail[n \ge N_1]{s_n - S}\) completely contained in the half-open range \([0,c/2)\text{.}\) But we not only assume that \(\sum a_k\) converges, we assume it converges absolutely. That is, \(\sum \abs{a_k}\) converges. Let \(\nseq{r}\) be the sequence of “remainders” for this convergent sequence:
\begin{equation*}
r_n = \sum_{k = n + 1}^\infty \abs{a_k} \text{.}
\end{equation*}
Fact 25.2.15 tells us that
\(\nseq{r} \to 0\text{,}\) so there exists some tail
\(\nseqtail[N_2]{r}\) also completely contained in the half-open range
\([0,c/2)\text{.}\) (Note that
\(\nseq{r}\)) is a sequence of non-negative terms because a convergent series of non-negative terms must have a non-negative sum value.)
Now let \(N\) be the greater of \(N_1\) and \(N_2\text{,}\) so that both the tails \(\absseqtail{s_n - S}\) and \(\nseqtail{r}\) are completely contained in \([0,c/2)\text{.}\) In particular, the first terms of these tails are both contained in this range:
\begin{align}
\abs{s_N - S} \amp \lt c/2 \text{,} \amp r_N \amp \lt c/2 \text{.}\tag{†}
\end{align}
Remember that the series we would like to investigate is \(\sum a_{\varphi(k)}\text{,}\) but the terms of this sequence are all mixed up. However, from the original series \(\sum a_k\text{,}\) the specific partial sum of the original series
\begin{equation*}
s_N = \sum_{k = 0}^N a_k = a_0 + a_1 + a_2 + \dotsb + a_N
\end{equation*}
involves only a finite number of terms, and all of these terms must appear eventually in the mixed-up sequence \(\seq{a_{\varphi(n)}}\text{.}\) Choose \(M\) large enough that this happens. (For example, we could choose \(M\) to be the largest index so that the “shuffled” index \(\varphi(M)\) is between \(0\) and \(N\text{,}\) inclusive.) The two partial sums \(s_N\) and \(t_M\) involve only a finite number of terms each, and so we can subtract them and manipulate them without worrying about convergence. But we have chosen \(M\) so that in the partial sum
\begin{equation*}
t_M = \sum_{k = 0}^M a_{\varphi(k)} \text{,}
\end{equation*}
all the terms of \(s_N\) appear, just in a mix-up order. So if we subtract \(t_M - s_N\text{,}\) all those common terms will cancel and we will only be left will those terms from \(t_M\) that did not appear in \(s_N\text{.}\) Because \(s_N\) involves all terms up to and including \(a_N\text{,}\) the terms remaining must all have indices that are larger than \(N\) (as part of the original sequence of terms \(\nseq{a}\)):
\begin{equation*}
\abs{t_M - s_N} = \abs{a_{j_1} + a_{j_2} + \dotsb + a_{j_\ell}}
\end{equation*}
for some collection of indices \(j_1,\dotsc,j_\ell\) that are all greater than \(N\text{.}\) Using the extended Triangle Inequality, we can say that
\begin{gather}
\abs{t_M - s_N} \le \abs{a_{j_1}} + \abs{a_{j_2}} + \dotsb + \abs{a_{j_\ell}}\text{.}\tag{††}
\end{gather}
In this form, because the indices are all larger than \(N\text{,}\) each of these absolute terms appears in the \(\nth[N]\) absolute remainder
\begin{equation*}
r_N = \sum_{k = N + 1}^\infty \abs{a_k} \text{.}
\end{equation*}
We know the series defining
\(r_N\) converges, and since it is an
infinite sum of
non-negative terms, it must add up to a larger value than the finite sum in
(††). So we can say
\begin{equation*}
\abs{t_M - s_N} \le r_N \text{.}
\end{equation*}
Now the partial sum \(t_M\) already contains all of the terms from \(s_N\) (and likely more), so if we go even further out into the partial sums \(\nseq{t}\) that will still be true. In other words, all of the analysis above is actually true for every difference \(\abs{t_n - s_N}\) for \(n \ge M\text{,}\) so that for such \(n\) it is always true that \(\abs{t_n - s_N} \le r_N\text{.}\)
We’re now ready to confirm that the tail \(\absseqtail[n \ge M]{t_n - S}\) is completely contained in the range \([0,c)\text{.}\) Because of the absolute values, we know that the terms of this tail will all be greater than or equal to zero; we just need to confirm that they are all less than \(c\text{.}\) So consider:
\begin{align*}
\abs{t_n - S} \amp = \abs{t_n - s_N + s_N - S} \\
\amp \le \abs{t_n - s_N} + \abs{s_N - S} \\
\amp \le r_N + \abs{s_N - S} \\
\amp \lt \frac{c}{2} + \frac{c}{2} \\
\amp = c \text{,}
\end{align*}
as required, where the first inequality above is the Triangle Inequality, the second is from our analysis of the terms of
\(\nseqtail[M]{t}\) compared to the specific partial sum
\(s_N\text{,}\) and the third is from
(†).
