If $\sum\limits_na_n$ converges, $f$ is a bijection and $|f(n)-n|<X$ for every $n$, for some fixed $X$, then $\sum\limits_na_{f(n)}$ converges > Let $f(n)$ be a bijection from $\Bbb N$ to $\Bbb N$ such that $$|f(n)-n|\lt X$$ for some fixed natural $X$ and $\forall n \in \Bbb N$. If $$\sum_{n=1}^\infty a_n$$ is convergent then prove that $$\sum_{n=1}^\infty a_{f(n)}$$ is also convergent. I don't know how to proceed with this problem. I am not able to see any direct way to use the cauchy criterion or any other method to prove that the given series is convergent using the fact that $$\sum_{n=1}^\infty a_{n}$$ is convergent and the condition on $f(n)$. Any hints will be helpful.

The basic idea is that if you sum $a_n$ or $a_{f(n)}$ up to some $N\gg X$, you get more or less the same terms except for about $2X$ terms near $a_N$, which are all very small.

We have $$\sum_{n=1}^N a_n - \sum_{n=1}^N a_{f(n)} =\sum_{j\ge N-X: f(j)>N} a_j - \sum_{N+X\ge j>N: f(j)< N} a_j $$ As $\sum_{n=0}^\infty a_n$ converges, $a_n\to 0$, so the $2X$ terms on the right hand side also tend to $0$ as $N\to\infty$. It follows that $\sum_{n=1}^\infty a_{f(n)}$ is convergent.