We know that if a series is convergent, then we can perform grouping on the series and the resulting series would still be convergent.
However,for an arbitrary series, grouping may not always give the same result. E.g. $a_n=(-1)^{n+1}$ is the non-convergent series
$1+(-1)+1+(-1)+...$
which can be grouped to
$(1-1)+(1-1)+...$ which converges to 0.
So my question is, in this link Establish convergence of the series: $1-\frac{1}{2}-\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}-...$ the solutions have grouped the terms in the series (before knowing whether the series converges or not) and showed it is convergent. How is that possible?
Suppose that $s_n=\sum\limits_{k=1}^n a_k$ denotes the $n$-th partial sum. Convergence of the series is equivalent to convergence of the sequence $(s_n)$.
By grouping some term you go to a subsequence $s_{n_k}$. Of course, if $s_{n_k}$ converges, that does not imply in general that so does $s_n$.
However, in the linked post the grouping is made in special way: The answerer took a block of consecutive positive terms, then the next block consists only of negative terms, etc. In this way you get that for $$n_k \le n \le n_{k+1}$$ you have that $s_n$ is between $s_{n_k}$ and $s_{n_{k+1}}$. (Since the sequence of partial sums is monotone on these intervals.)
So in this case if you get that $\lim\limits_{k\to\infty} s_{n_k}=\ell$, you immediately see that the limit of the sequence $(s_n)$ is the same, i.e. $\lim\limits_{n\to\infty} s_n=\ell$.