convergent sequence / prove of reordering rule Could someone show me please how to solve the following question?
Prove/show or disprove the following reordering rule:
If $(a_n)_{n \in \mathbb{N}}$ is a convergent sequence and $\pi : \mathbb{N} \to \mathbb{N}$ is a bijective function, then 
$\lim_{n \to \infty} a_n = \lim_{n \to \infty} a_{\pi(n)}$
 A: Hint:
If $\pi$ is a bijection, what is the following limit?
$$
\lim_{n\rightarrow\infty}\pi(n)
$$
You can show that it must be infinite.  Because $(a_n)$ is convergent, for any $\epsilon$ you can find an $N$ such that $\lvert a_n-a\rvert<\epsilon$ for all $n\geq N$, where $a=\lim a_n$. Now, can you somehow find an $M$ such that $\pi(n)\geq N$ whenever $n\geq M$?
A: Define $a=\lim{n\to\infty}a_n$.
Be $\epsilon > 0$. Since $(a_n)$ is convergent, there exists an $N\in\mathbb N$ so that $\left|a_n-a\right|<\epsilon$ for all $n>N$. Since there are only finitely many natural numbers $\le N$, there exists $N' = \max\{\pi^{-1}(n):n\le N\}$. Then for every $n>N'$, we have $\pi(n)>N$, and therefore $\left|a_{\pi(n)}-a\right|<\epsilon$. Thus $(a_{\pi(n)})$ also converges to $a$.
A: I think this assertion is correct. A proof is:
For $(a_n)_{n\in\mathbb{N}}$ and $(a_{\pi(n)})_{n\in\mathbb{N}}$, since $\pi$ is bijective, we can conclude that 
$\varlimsup_{n} a_n=\varlimsup_{n} a_{\pi(n)}$ and also $\varliminf_{n} a_n=\varliminf_{n} a_{\pi(n)}$. 
We already know that $\varlimsup_{n} a_n=\varliminf_{n} a_n$, so $\varlimsup_{n} a_{\pi(n)}=\varliminf_{n} a_{\pi(n)}$. 
Conclusion follows.
