A trivial question on the sequences $\{a_n\}$ and $\{a_{n+1}\}$ converging to the same limit using the definition of convergence only

Question

Let's say I didn't know that if a sequence $\{a_n\}$ converges to some number $a$, then all subsequences of $\{a_n\}$ also converge to $a$.

I want to show using the definition of convergence that if $\{a_n\}$ converges to $a$, then $\{a_{n+1}\}$ converges to $a$.

Since $\{a_n\}$ we have for all $\epsilon > 0$, there exists a natural number $N$ such that for all $n \in \mathbb{N}$ $$n \geq N \implies |a_n - a | < \epsilon$$ I'm aware that some books use $n > N$, but this one uses $n\geq N$. Since it doesn't matter, let's stick with $n\geq N$ to be consistent =)

Clearly $n+1 > n \geq N$, so I should be able to deduce from the above implication that $$|a_{n+1} - a| < \epsilon$$ But I have some doubts. My main question is if I pick any index $m > n$, will I be able to immediately conclude that $$|a_m - a| < \epsilon$$ In other words, picking an index m > n allows me to "update" the index in $|a_n - a| < \epsilon$ to $|a_m - a| < \epsilon$?

Answer 1

If $(a_n)_{n\ge 0}$ converges to $a$ then

for each $\epsilon>0$ there exist $N\in \Bbb N$ such that

$$n\ge N \implies |a_n-a|<\epsilon$$

thus

$$m>n\ge N \implies |a_m-a|<\epsilon$$