If the sequence $(a_n) $ converges, then $(a_{n-1}) $ also converges to the same limit as $(a_n) $ If the sequence $(a_n)_{n \in \mathbb {N}} = (a_n)$ converges to the real number $L $, then the sequence $(a_{n-1})_{n \in \mathbb {N}} = (a_{n-1}) $ also converges to $L $.
This appears true intuitively, but I'm trying to write a proof.
Suppose $(a_n) $ converges to $L $. Then for every $\epsilon > 0$, there is an $N \leq n $ such that $a_n \rightarrow L $. Choose $N=n-2$. Then $N < n-1$, and since the chosen $\epsilon $ corresponds with the selected $N $ (in reference to the order of the quantifiers in the definition), it follows that $a_{n-1} \rightarrow L $.
Is this sort of correct? Is there a clearer way using the definition of the limit of a sequence?
 A: First of all your definition of convergence is kind of off:
$(a_n)$ converges to $L$ if and only if for all $\epsilon>0$ there is an $N \in \mathbb{N}$ such that $n>N$ implies $|a_n-L|<\epsilon$.
Now suppose $(a_n)$ converges to $L$. Fix $\epsilon>0$. Then there is an $N \in \mathbb{N}$ such that $n>N$ implies $|a_n-L|< \epsilon$. Then for this same $N$, $n-1>N$ implies $|a_{n-1}-L|<\epsilon$, hence $(a_{n-1}) \to L$.
A: I don't think you've used the definition of the convergence of a sequence correctly, and in fact I'm not sure I understand what you mean by this:
"Then for every $\epsilon > 0$ there is an $N\leq n$ such that $a_n\rightarrow L$"
I think what you mean to say is the for any $\epsilon >0$, there is an $N$ such that if $n \geq N$, then $|a_n - L|<\epsilon$.
A: Definition: the $N$-th tail of a sequence is obtained by dropping the first $(N-1)$ terms of the sequence.
You should prove the following statement, which is more general, but at the same time much simpler, than your problem: 
A sequence converges to $L$ if and only if its every tail converges to $L$.
