Let $\sum a_n=s$ be conditional convergent, $\sum a_{f(n)}=t\neq s$. Show $\forall\ N, \exists\ n$, such that $|n-f(n)|>N$.
Here, $f: \Bbb N\to\Bbb N$ is a bijection.
How to prove?
My attempt. Let the partial sums of $\sum a_n$ and $\sum a_{f(n)}$ be $s_n,t_n$. Argue by contradiction. If $\exists\ M>0$, such that for any $n$, $|n-f(n)|\leq M$. Then for $n>M$, $a_1,\cdots,a_{n-([M]+1)}$ appears in $t_n$. Then how to estimate $|s_n-t_n|$?