This is a very interesting observation. But the Floor magic is useful and I am quite sure I used it several years ago without any problem.

The series does not absolutely converge. Therefore, such a rearrangement is not grounded (see Wiki for info).

@user64494 there is no relevant rearrangement, please look again. What you are referring to are rearrangements of non-absolutely convergent sums where the rearrangement extracts sub-series at different rates, which is not the case here.

(1) This was no rearrangement, just an application of associativity of addition. (2) Exercise: show that "bounded rearrangements' do not change the value or convergence/divergence. By this term I mean reorderings wherein, for some finite m, all elements move no more than m places from where they started.

@Roman: More exactly speaking, you use grouping which is not grounded. A simple example: a series 1-1+1-1+1... diverges, but (1-1)+(1-1)+... converges. Hope I am clear.

@user64494 The original series is convergent. Therefore it has nothing to do with your "counterexample". For convergent series regrouping does not change the result (the absolute convergence is not required). So Roman is correct.

@user64494 If $g_0, g_1, g_2, g_3, \dots$ is the sequence of partial sums of the "grouped" sequence, then the sequence of partial sums of the original also includes terms like $g_n+\frac1{4n+1}$, $g_n+\frac1{4n+1}+\frac1{4n+2}$, and $g_n+\frac1{4n+1}+\frac1{4n+2}-\frac1{4n+3}$. These extra 1-3 terms approach $0$ as $n\to\infty$ (unlike what happens in the $1-1+1-1+1\dots$ example), so grouping cannot change the convergence.

@drer: Can you ground " For convergent series regrouping does not change the result (the absolute convergence is not required)"?

@MishLavrov: Don't understand you at all. Can you refer to some theorems?

@MishaLavrov I have to agree with the comment that your convergence argument is suspect, or at least missing some detail. You show that a finite extension of the SoPS converges to zero but that does not imply the full tail does so.

@user64494 The argument that I meant is exactly the argument you included in your latest comment, in somewhat different words.

I've removed non technical aspects of this otherwise useful conversation, let's keep it this way.

@user64494 If a sequence $\{a_n\}$ is convergent, then any its subsequence $\{a_{n_k}\}$ is convergent as well and has the same limit as $\{a_n\}$. This is a well-known theorem, which you certainly can easily find.

@drer: But Roman proves the convergence only for a special subsequence of the sequence of the partial sum. Hope you feel the difference.

@user64494 I have stated already in question that the original series is convergent. This can be easily proved by Dirichlet's test.

@drer: Sorry, this is off-topic. You avoid of an answer.

@user64494 What answer do you mean?

@drer: An answer to "But Roman proves the convergence only for a special subsequence of the sequence of the partial sum. Hope you feel the difference".