I was attempting to understand an epsilon n proof when this was the stated chain of implications:
$$\forall \epsilon > 0, \exists N_2 > 1 : n + 1 \geq N_2 \implies | c_n - L | < \epsilon$$ Then: $$\forall \epsilon > 0, \exists N_2 > 1 : n \geq N_2 \implies | c_{n-1} - L | < \epsilon$$
Is this step "mathematically sound", and if so, is there some possible intuition behind this that could aid in my understanding?
Many thanks!
Yes, that is correct. Take $\varepsilon>0$. You are assuming that there is some $N_2\in\Bbb N\setminus\{1\}$ such that $n+1\geqslant N_2\implies|c_n-L|<\varepsilon$. But then, if $n\geqslant N_2$, you have $(n-1)+1\geqslant N_2$, and therefore $|c_{n-1}-L|<\varepsilon$.
$$n + 1 \geq N_2 \implies | c_n - L | < \epsilon$$
says that $| c_n - L | < \epsilon $ is true for all $n \geq N_2-1.$
which is the same as saying that $| c_{n-1} - L | < \epsilon $ is true for all $n \geq N_2.$
To convince yourself, replace $N_2$ with a positive integer.
