Rearrangement of a Different Convergence (Example 3.53)

Walter Rudin Principles of Mathematical Analysis, chapter 3, example 53:

Consider the convergent series $$1-1/2+1/3-1/4+...$$ and one of its rearrangements $$1+1/3-1/2+1/5+1/7-1/4+1/9+1/11-1/6+...$$ As we have seen, the rearrangement $s_n$ if it converges does so at a point larger than the point of convergence for the original series. Prove $s_n$ converges.

Proof: For any sequence $a_n$, let $f~:~ℕ→ℕ$ be such that $a_{f(n)}$ and $a_n-a_{f(n)}$ are convergent sequences. Then we see $(\text{lim }a_n-a_{f(n)})+(\text{lim }a_{f(n)})=\text{lim }a_n$ so that $a_n$ converges. As it applies to this problem, we will show the subsequence $s_{3n}$ converges, and since $\text{lim }s_n-s_{f(n)}=0$ where $f(n)$ rounds $n$ to the next highest multiple of $3$, $s_n$ converges (technically $s_{f(n)}$ has different terms from $s_{3n}$, but $\text{lim }s_{f(n)}=\text{lim }s_{3n}$ since the former merely has finite copies of each of the latter's terms).

It suffices to compare the sum's terms $a_n$ to $\dfrac{4}{n^2}$, which forms a convergent series: $$|a_n|=a_n=\dfrac{1}{4n-3}+\dfrac{1}{4n-1}-\dfrac{1}{2n}=\dfrac{8n-3}{2n(4n-3)(4n-1)}$$ $$≤\dfrac{4}{(4n-3)(4n-1)}≤\dfrac{4}{n^2}~~\square$$