Here is Prob. 1, Sec. 3.7, in the book Introduction to Real Analysis by Robert G. Bartle and Donald R. Sherbert, 4th edition:
Let $\sum a_n$ be a given series and let $\sum b_n$ be the series in which the terms are the same and in the same order as in $\sum a_n$ except that the terms for which $a_n = 0$ have been omitted. Show that $\sum a_n$ converges to $A$ if and only if $\sum b_n$ converges to $A$.
My Attempt:
In order to have a series at all for $\sum b_n$, we must of course assume that there are non-zero terms in $\sum a_n$; rather, we must also assume that there are infinitely many non-zero terms of the series $\sum a_n$, for otherwise both $\sum_{n=1}^\infty a_n$ and $\sum_{n=1}^\infty b_n$ be finite sums having the same non-zero terms and will thus be equal.
Am I right?
For the sake of definiteness, let us assume that $a_n = 0$ whenever $n$ is a prime number and that $a_n$ is non-zero otherwise. Let $\left( s_n \right)_{n \in \mathbb{N}}$ and $\left( t_n \right)_{n \in \mathbb{N}}$ be the sequences of the partial sums of $\sum a_n$ and $\sum b_n$, respectively. Then we note that $$ \begin{align} t_1 &= s_1 = s_2 = s_3, \\ t_2 &= s_4 = s_5, \\ t_3 &= s_6 = s_7, \\ t_4 &= s_8, \\ t_5 &= s_9, \\ t_6 &= s_{10} = s_{11}, \\ t_7 &= s_{12} = s_{13}, \\ t_8 &= s_{14}, \\ t_9 &= s_{15}, \\ t_{10} &= s_{16} = s_{17}, \end{align} $$ and so on and so forth. In this way, we observe that $\left( t_n \right)_{n \in \mathbb{N}}$ is a subsequence of $\left( s_n \right)_{n \in \mathbb{N}}$ and therefore the former sequence has the same limit as the latter.
Is this kind of "proof" satisfactory enough, I wonder? I would like to make it more general and rigorous though.
Reading off the above array from the right to left, we observe that the terms of the sequence $\left( s_n \right)_{n \in \mathbb{N}}$ also appear in the sequence $\left( t_n \right)_{n \in \mathbb{N}}$ and the predecessor-successor relationship between terms is preserved in passing from the latter sequence to the former. Thus (intuitively at least) the former sequence cannot have any limit different from the limit of the latter sequence.
How to make this part of my argument more precise and rigorous, employing only what has preceded this particular exercise problem in the book?
In general, how to give a proof of each one of the two parts in the full level of generality, rigor, and detail?
Let $\sum a_n$ and $\sum b_n$ be two series satisfying the hypothesis
Let $s_n = a_1 + a_2 + a_3 + \cdots +a_n $, and let $t_n = b_1+ b_2 + b_3 + \cdots + b_n $ be the sequences of the partial sums of $\sum a_n$ and $\sum b_n$, respectively.
Suppose there are $m$ terms $a_i$ equal to 0 in the partial sum of $a_n$
Thus, $n-m$ non-zero terms are there in the sum $s_n$ , which is exactly the first $n-m$ terms of $\sum b_n$.
Thus, $$s_n = t_{n-m}. $$
Therefore , if $s_n$ converges, then $t_n$ converges. And, conversely, if $t_n$ converges then $s_n$ converges.
Let $lim s_n = lim t_n = \alpha$, say.
Hence $\sum a_n$ converges to $\alpha$ $\iff$ $\sum b_n$ converges to $\alpha$.