# Prove that the series still converges under a permutation

Let $f$ be a permutation of positive integers such that for all $n$ we have $|f(n)-n|<2014$. Prove that if $\sum^{\infty}_{n=1}a_n$ is a convergent series, then so is $$\sum^{\infty}_{n-1}a_{f(n)}$$.

By hypothesis, there is a number $L$ such that $\forall \epsilon, \exists N,$ such that $n\geq N$ implies that $|\Sigma_{i=1}^{n}a_{i}-L|<\epsilon$. Since the permutation moves a number within a certain range, there is an $N'$ such $n\geq N'$ implies that $\{a_i|1\leq i \leq N\}\subset\{a_i|1\leq i\leq n\}$. I don't know what I can do next.

Hint: First, since $\sum a_n$ is convergent, we must have $a_n \to 0$ as $n\to \infty$. So, given $\epsilon > 0$, there exists $N_1$ such that if $n\geq N_1$ then $\lvert a_n \rvert < (\epsilon/2) / 2014 = \epsilon/4028$.

Also, as you pointed out, there exists $N_2$ such that if $N \geq N_2$ then $\lvert (\sum_{n=1}^{N} a_n) - L\rvert < \epsilon /2$.

Now let $b_n = a_{f(n)}$, and suppose $N \geq 2014 + \max(N_1, N_2)$. We have $$\Big\lvert \sum_{n=1}^N b_n - L \Big\rvert = \Big\lvert \sum_{n=1}^N a_{f(n)} - L \Big\rvert = \Big\lvert \sum_{n\in S} a_{n} - L \Big\rvert,$$ where $S= \{f(x) \mid x = 1, 2, \ldots, N\}$. Now, the conditions on $f$ guarantee that $\{ 1, 2, \ldots, N-2014\} \subseteq S$ (prove this!), and therefore the set difference $S - \{ 1, 2, \ldots, N-2014\}$ has cardinality $N - (N-2014) = 2014$, and its elements are all greater than or equal to $N-2013 \geq N_1$.

The desired conclusion now follows, by splitting $S$ into two subsets and using standard tricks of analysis.

Hint:

If $|f(n) - n | \le \delta$ for all $n$ then for all $n$ we have \begin{eqnarray} f\,(\{1, 2, \ldots, n\})&\subset &\{ 1, 2, \ldots, n+\delta\} \end{eqnarray} Moreover, if $f$ is also surjective then \begin{eqnarray} \{1, 2, \ldots n-\delta\} &\subset & f\,(\{1, 2, \ldots, n\}) \end{eqnarray} In conclusion: if $f$ is bijective and $|f(n)-n|\le \delta$ for all $n$ then we have $$f\,(\{1, 2, \ldots, n\})\Delta \{ 1, 2, \ldots, n\}\subset \{ n-\delta, \ldots, n + \delta \}$$ and therefore $$\left|\sum_{i=1}^n a_{f(i)} - \sum_{i=1}^n a_{i} \right| \le \sum_{i=n-\delta}^{n+\delta} |a_{i}|\to 0$$ as $n \to \infty$

• @becko: If the series is convergent, but $\sum_n |a_n|$ is divergent ( like for instance $a_n = \frac{(-1)^{n-1}}{n}$ for $n\ge 1$) then for every $S \in [\infty, \infty]$ there exists a permutation $\sigma$ so that $\sum_n a_{\sigma(n)} = S$. Moreover, one can also choose permutations so that the new series does not have a sum. This is a theorem of Riemann, not that hard to prove. Commented Oct 7, 2017 at 15:28