# Convergence sequences also always bounded?

So this is the question;

Let $(a_n)_{n=1}^{\infty}$ be a sequence in $\mathbb{R}$. How could you prove the following statement.

If $(a_n)_{n=1}^{\infty}$ converges to $0$ then $(a_n)_{n=1}^{\infty}$ is bounded.

$$\lim_{n\to\infty}a_n=L\;,\;\;L\in\Bbb R\implies \exists\;N\in\Bbb N\;\;\text{such that}$$
$$n>N\implies |a_n-L|<1\iff L-1<a_n<L+1\implies$$
$$\forall\,n\in\Bbb N\;\,\;|a_n|\le max\left\{\;|a_1|, |a_2|,...,|a_N|, L+1\;\right\}$$
As $a_n$ converges to $0$, there exists a natural number $N$ such that for all $n>N$, $|a_n|<1$ (where I arbitrarily decided to pick $1$ here).
So all the terms bigger than $N$ are bounded by $1$. Now there are only finitely many terms before $N$. Let $K = max\{a_1, \ldots, a_N\}$. Then the entire sequence is bounded by $\max\{K,1\}$.