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 23.3.31, we will instead verify that
\(\seq{t_n - S} \to 0 \text{.}\) Following
Corollary 23.3.32, 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 26.2.15 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{equation}
\abs{t_M - s_N} \le \abs{a_{j_1}} + \abs{a_{j_2}} + \dotsb + \abs{a_{j_\ell}}\text{.}\tag{††}
\end{equation}
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
(†).
