# Proof that Convergent Sequences of Real Numbers are Bounded

We will now look at an extremely important result regarding sequences that says that if a sequence of real numbers is convergent, then that sequence must also be bounded.

Theorem: If $(a_n)$ is a sequence of real numbers that is convergent to $L \in \mathbb{R}$, that is $\lim_{n \to \infty} a_n = L$, then $(a_n)$ is also bounded. |

**Proof:**Suppose that $(a_n)$ is convergent to $L$, in other words, $\lim_{n \to \infty} a_n = L$. Let $\epsilon_0 = 1$. By the definition that $(a_n)$ is convergent to $L \in \mathbb{R}$ it follows that for $\epsilon_0 = 1$ there exists an $N \in \mathbb{N}$ such that if $n ≥ N$ then $\mid a_n - L \mid < \epsilon_0 = 1$. Equivalently, $-1 < a_n - L < 1$ and so $L - 1 < a_n < L + 1$ for $n ≥ N$.

- Thus for $n ≥ N$ we have that $\mid a_n \mid < \mid L \mid + 1$.

- Now there is only a finite number of terms to consider, namely the terms $a_1, a_2, ..., a_{N-1}$. Let $M = \mathrm{max} \{ \mid a_1 \mid, \mid a_2 \mid, ..., \mid a_{N-1} \mid, \mid L \mid + 1 \}$. Therefore for all $n \in \mathbb{N}$, $\mid a_n \mid < M$ and so $(a_n)$ is bounded. $\blacksquare$

It is important to recognize in the theorem above that we could not have just taken the maximum value from the start because we cannot take the maximum of an infinite set, only a finite set. From the definition of the limit of a sequence we had that for the infinitely many $n ≥ N$ that $\mid a_n \mid < \mid L \mid + 1$. Thus, if we take the largest value out of $\mid a_1 \mid, \mid a_2 \mid, ..., \mid a_{N-1} \mid, \mid L \mid + 1$ then we are guaranteed to find a bound for our sequence.

It is also important to note that the converse of this theorem is not always true, that is just because a sequence $(a_n)$ is bounded does NOT imply $(a_n)$ to be convergent. For example, the sequence $((-1)^n) = (-1, 1, -1, 1, -1, ...)$ is bounded since $\mid a_n \mid ≤ 1$ for all $n \in \mathbb{N}$, however, clearly $((-1)^n$ does not converge.