Here is Theorem 8.14 in the book Mathematical Analysis - A Modern Approach to Advanced Calculus by Tom M. Apostol, 2nd edition:
Let $\sum a_n$, $\sum b_n$ be related as in Definition 8.12. Assume that there exists a constant $M > 0$ such that $p(n+1) - p(n) < M$ for all $n$, and assume that $\lim_{n \to \infty} a_n = 0$. Then $\sum a_n$ converges if, and only if, $\sum b_n$ converges, in which case they have the same sum.
Now here is Definition 8.12:
Let $p$ be a function whose domain is the set of positive integers and whose range is a subset of the set of positive integers such that$$ i) \qquad \qquad p(n) < p(m), \qquad \mbox{ if } n < m. $$Let $\sum a_n$ and $\sum b_n$ be two series related as follows:$$b_1 = a_1 + \cdots + a_{p(1)} $$$$ii) \qquad \qquad b_{n+1} = a_{p(n)+1} + a_{p(n)+2} + \cdots + a_{p(n+1)} \qquad \mbox{ if } n = 1, 2, 3, \ldots. $$Then we say that $\sum b_n$ is obtained from $\sum a_n$ by inserting parentheses, and that $\sum a_n$ is obtained from $\sum b_n$ by removing parentheses.
And, here is Theorem 8.13:
If $\sum a_n$ converges to $s$, every series $\sum b_n$ obtained from $\sum a_n$ by inserting parentheses also converges to $s$.
Finally, here is Apostol's proof of Theorem 8.14:
If $\sum a_n$ converges, the result follows from Theorem 8.13. [So far so good!] The whole difficulty lies in the converse deduction. [Indeed!] Let$$s_n = a_1 + \cdots + a_n, \qquad t_n = b_1 + \cdots + b_n, \qquad t = \lim_{n \to \infty} t_n. $$Let $\varepsilon > 0$ be given and choose $N$ so that $n > N$ implies$$\left\lvert t_n - t \right\rvert < \frac{\varepsilon}{2} \qquad \mbox{ and } \qquad \left\lvert a_n \right\rvert < \frac{\varepsilon}{2M}. $$[OK!] If $n > p(N)$, we can find $m \geq N$ so that $N \leq p(m) \leq n < p(m+1)$. [Why? How? How can we show this to be true?] For such $n$ we have$$\begin{align} s_n &= a_1 + \cdots + a_{ p(m+1)} - \left( a_{n+1} + a_{n+2} + \cdots + a_{ p(m+1) } \right) \\&= t_{n+1} - \left( a_{n+1 } + a_{n+2 } + \cdots + a_{ p(m+1) } \right),\end{align}$$and hence$$\begin{align}\left\lvert s_n - t \right\rvert &\leq \left\lvert t_{n+1} - t \right\rvert + \left\lvert a_{n+1} + a_{n+2} + \cdots + a_{p(m+1)} \right\rvert \\&\leq \left\lvert t_{n+1} - t \right\rvert + \left\lvert a_{p(m)+1} \right\rvert + \left\lvert a_{p(m)+2} \right\rvert + \cdots + \left\lvert a_{p(m+1)} \right\rvert \\ &< \frac{\varepsilon}{2} + \big( p(m+1) - p(m) \big) \frac{\varepsilon}{2M} = \frac{\varepsilon}{2} + \frac{\varepsilon}{2} = \varepsilon.\end{align}$$This proves that $\lim_{n \to \infty} s_n = t$.
I think I'm clear on each and every detail in the above proof except for the spot therein that I have indicated.
In particular, how to show that the following holds?
Let $n$ be any positive integer $> p(N)$. Then there exists a positive integer $m \geq N$ such that $N \leq p(m) \leq n < p(m+1)$.