Why does $\lim_{n \to\infty}a_{n+1} = \lim_{n\to\infty}a_n$? Assume that $\{a_n\}$ is a convergent sequence. How to use the definition of a limit of a sequence to prove that $\lim\limits_{n \to\infty}a_{n+1} = \lim\limits_{n\to\infty}a_n$?
Write $L=\lim_{n\to\infty}a_n$. This means that for every $\epsilon&gt;0$ there exists a whole number $N$ such that for all integers $n\ge N$, $$|L-a_n|&lt;\varepsilon$$ Now, to prove $\lim_{n\to\infty}a_{n+1}=L$ as well, consider any $n\ge N$ and note that $n+1\ge N$. So by our initial statement, $|L-a_{n+1}|&lt;\varepsilon$.
I think this is just a matter of notation.

For $n \ge 1$,
considering $a_{n+1}$ means the sequence $a_2, a_3, a_4, a_5, \dots$, and
considering $a_n$ means the sequence $a_1, a_2, a_3, a_4, a_5, \dots$.

Limits are a matter of the eventual behaviour of the sequence, and it is not surprising that the limit (if it exists) is the same for "both" — it's the same sequence whose eventual behaviour you're considering.

Put differently, $$\lim_{n\to\infty} a_{n+1} = \lim_{m-1\to\infty} a_m = \lim_{m\to\infty} a_m = \lim_{n\to\infty} a_n$$
as $m-1\to\infty$ and $m\to\infty$ are the same thing.