Rearrangements that never change the value of a sum Which bijections $f:\{1,2,3,\ldots\}\to\{1,2,3,\ldots\}$ have the property that for every sequence $\{a_n\}_{n=1}^\infty$,
$$
\lim_{n\to\infty} \sum_{k=1}^n a_k = \lim_{n\to\infty} \sum_{k=1}^n a_{f(k)},
$$
where "$=$" is construed as meaning that if either limit exists then so does the other and in that case then they are equal?
It is clear that there are uncountably many of these.
Might it just be that $\{f(n)/n : n=1,2,3,\ldots\}$ is bounded away from both $0$ and $\infty$?
These bijections form a group.  Can anything of interest be said about them as a group?
PS: Here's another moderately wild guess (the one above appears to be wrong): Might it be just the bijections whose every orbit is finite?
 A: The hypothesis is wrong.
Let $f$ be the following sequence:

$1,2 ; 3,5,4,6; 7,9,11,8,10,12; \cdots$

where each block has two more numbers than the previous block and goes through the odd numbers in order before the even ones.
Clearly $f(n) \in n + O(\sqrt{n})$ as $n \to \infty$.
Now consider the following series:

$\frac{1}{1} - \frac{1}{1} \quad + \quad \frac{1}{2} - \frac{1}{2} + \frac{1}{2} - \frac{1}{2} \quad + \quad \frac{1}{3} - \frac{1}{3} + \frac{1}{3} - \frac{1}{3} + \frac{1}{3} - \frac{1}{3} \quad + \quad \cdots$

split into blocks in the same manner where each block has an alternating sum of the same reciprocal of the block index. Clearly this series converges to $0$. However after applying $f$ to the sequence in the series we get:

$\frac{1}{1} - \frac{1}{1} \quad + \quad \frac{1}{2} + \frac{1}{2} - \frac{1}{2} - \frac{1}{2} \quad + \quad \frac{1}{3} + \frac{1}{3} + \frac{1}{3} - \frac{1}{3} - \frac{1}{3} - \frac{1}{3} \quad + \quad \cdots$

which clearly does not converge.
Also, it is easy to see that we can insert arbitrarily large blocks of identity function between the blocks in the permutation, and corresponding blocks of zeros in the series, and the behaviour is exactly the same. This means that there is a counterexample where $f(n) \in n + o(n)$ as $n \to \infty$, equivalently where $\lim_{n\to\infty} \frac{f(n)}{n}$ exists.
A: My guessed interpretation of the second hypothesis is wrong too. (The original second hypothesis was already invalided by the first counter-example.)
Let $f$ be the following sequence:

$1;3,2;5,6,4;8,9,10,7;\cdots$

where each block has one more number than the previous block and is essentially a cyclic shift of elements at those positions. So $f$ has infinite order.
Then rearranging a series using $f$ does not change the existence and value of its limit because the magnitude of the discrepancy between partial sums is always at most the sum of magnitude of two terms with eventually increasing indices. If one converges, the other must hence converge since the terms go individually to zero.
A: Firstly I shall prove that indeed the rearrangements $R$ that preserve the convergence and value of every series form a group.$\def\nn{\mathbb{N}}$ Every rearrangement can of course be considered as a permutation on $\nn$. Every sequence is also just a function on $\nn$. We shall use $\sum^\infty x$ to denote the limit of the partial sums of $x$ if it exists and $null$ otherwise.
For any $f \in R$:
  $\sum^\infty a = \sum^\infty ( a \circ f )$ for any sequence $a$:
  For any sequence $a$:
    $\sum^\infty ( a \circ f^{-1} ) = \sum^\infty ( a \circ f^{-1} \circ f ) = \sum^\infty a$.
  Therefore $f^{-1} \in R$.
For any $f,g \in R$:
  $\sum^\infty a = \sum^\infty ( a \circ f )$ for any sequence $a$:
  $\sum^\infty a = \sum^\infty ( a \circ g )$ for any sequence $a$:
  For any sequence $a$:
    $\sum^\infty a = \sum^\infty ( a \circ f ) = \sum^\infty ( a \circ f \circ g )$.
  Therefore $f \circ g \in R$.
[Okay I did look at the paper (and I hate pay-walls for mathematics papers) and the collection that they say is not a group is the permutations that make a convergent series into another convergent one. The paper even mentions that the collection you are interested in is obviously a group.]
A sufficient condition for the rearrangement sequence to be sum-preserving is that there is a natural $k$ such that for any natural $n$ no more than $k$ numbers in $[1..n]$ are missing from the length-$n$ prefix of the sequence.
Clearly if the above condition holds, then the discrepancy between the partial sums of the original and rearranged series are bounded by the sum of magnitude of at most $2k$ terms. These terms have eventually increasing indices, because the numbers in $[1..n]$ that are not in the length-$n$ prefix will eventually appear in a longer prefix, and the numbers in the length-$n$ prefix that are not in $[1..n]$ are in $[1..m]$ for some $m > n$. Since the terms eventually go to zero, the sum of any $2k$ terms with eventually increasing indices must also go to zero.
This condition is far from necessary, because the following permutation is sum-preserving but has $\limsup_{n\to\infty} \frac{f(n)}{n} = \infty$ and $\liminf_{n\to\infty} \frac{f(n)}{n} = 0$:

$1;3,2;9,8,7,6,5,4;\cdots$

where the $k$-th block has $k!$ elements and is simply reversed. The reason is that for any $ε > 0$ there is some natural $k$ such that all the partial sums of any block beyond the $k$-th have magnitude less than $ε$, and hence also the partial sums of the block's reverse. This means that the rearranged series has the same sum by Cauchy convergence.
