convergence divergence - 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.

Answer

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.