TITLE: Special case of Riemann Rearrangement Theorem
QUESTION [5 upvotes]: Let $\sum_{k=1}^\infty a_{\varphi(k)}$ be a rearrangement of a conditionally convergent series $\sum_{k=1}^\infty a_k$. Prove that if $\{\varphi(k)-k\}$ is a bounded sequence, then $\sum_{k=1}^\infty a_{\varphi(k)}=\sum_{k=1}^\infty a_k$.
I can't find the solution anywhere and I can't figure it out. Thanks for your help. My understanding of the problem is that if we limit the "space" between the difference of terms then we don't need to reach into an asymptote to find the term $a_{\varphi(k)}$ which creates the same effect as if we were dealing with finite sums instead.

REPLY [4 votes]: Suppose 
$$
\sum_{k=1}^\infty a_k = L,
$$
and suppose $\phi: \mathbb{N} \to \mathbb{N}$ is such that there is an $N$ for which
 $\lvert \phi(k)-k \rvert \leq N$ for all $k \in \mathbb{N}$.
Let $$S_n = \sum_{k=1}^n a_k$$
and 
$$S'_n = \sum_{k=1}^n a_{\phi(k)}.
$$
Now consider
$$
S'_{n+N} - S_n = \sum_{k=1}^{n+N} a_{\phi(k)} - \sum_{k=1}^n a_k.
$$
If $n$ is much larger than $N$,
 then all the terms $a_1$, $a_2$, …, $a_{n}$ appear in
$a_{\phi(1)}$, $a_{\phi(2)}$, …, $a_{\phi(n+N)}$, 
because $N-k \leq \phi(k) \leq N+k$.
Let 
$$A_n=\{\phi(1), \phi(2), \ldots, \phi(n+N) \} \setminus \{1, 2, \ldots, n\}
$$
be the set of all indices of terms of $S'_{n+N}$ that don't arise in this way. 
We have 
$$A_n \subseteq \{n, n+1, \ldots, n+N\}.$$
Now we have
$$S'_{n+N} - S_n = \sum_{k=1}^{n+N} a_{\phi(k)} - \sum_{k=1}^n a_k = \sum_{i \in A_n} a_i,$$
or in other words
$$S'_{n+N} = \sum_{i \in A_n} a_i + S_n.$$
By definition,
$$\lim_{n \to \infty} S_n = L,$$
while since the series converges
$$\lim_{n \to \infty} \sum_{i \in A_n} a_i = 0.
$$
Therefore 
$$\lim_{n \to \infty} S'_{n+N} = L.$$
Thus
$$
\sum_{k=1}^\infty a_{\phi(k)} = L,
$$
as desired.