Prove a limit property using formal definition Prove that for any $k \ \in \ \mathbb{N}$ if $a_{n+k}$ converges to $a$, then $a_{n}$ also converges to $a$.    
I can prove the other way round since $n+k > n$, but this is creating trouble.    
Any help will be appreciated.
Thanks.
 A: Suppose $a_{n+k}\to a$ and let $\varepsilon>0$. There is $N$ s.t. $$n\geq N\implies |a_{n+k}-a|<\varepsilon.$$
In particular, if $n\geq N+k$, then $|a_n-a|<\varepsilon$, and thus, the claim follow.
A: Let $\epsilon > 0$. We want to show that $a_n \to a$, given that for any $k$, we have $a_{n + k} \to a$. Let us put $k = 1$, say. Let us call $b_k = a_{k + 1}$, so that $b_k \to a$.
Now, since $b_k \to a$, for the given $\epsilon > 0$, there exists $N \in \mathbb N$ such that $n \geq N \implies |b_n - a| < \epsilon$, which can be rewritten as $|a_{n+1} - a| < \epsilon$.
Now, simply note that $n \geq N-1 \implies n+1 \geq N \implies |a_n - a| < \epsilon$. That is, for the given $\epsilon$, we found a constant $ = N - 1$ which satisfies the given conditions. Hence, it follows that $a_n \to a$.
Two remarks after completion:

We never used the fact that $a_{n+k}$ converges for all $k$! Indeed, the existence of one such finite $k$ is enough to deduce this theorem. It will follow if we just replace everywhere, the constant $1$ by $k$ in the proof above.
The limit of a sequence is a property concerning it's "tail", in some sense. That is, if you take a sequence, and affect only finitely many of it's terms, then it does not change whether the sequence converges or not. In our case, we are basically trimming off the first $k$ terms of the sequence $a_n$, and getting the sequence $a_{n+k}$. Note that this affects only the first $k$ terms of the sequence $a_n$,which are getting chopped off, so indeed both these sequences will either converge or diverge together, and that is indeed the case in this problem as well.

