Comparison Theorem
Let \(x_n\) and \(y_n\) be two convergent sequences with limits \(x\) and \(y\) respectively. If there exists \(N_0\) such that \(x_n \leq y_n\) for all \(n \geq N_0\), then \(x \leq y\).

Proof

Suppose for the sake of contradiction that \(x > y\). We are given that there exists an \(N_0 \in \mathbb{N}\) such that when \(n \geq N_0\), then

$$ \begin{align*} x_n \leq y_n. \end{align*} $$

Now, Take \(\epsilon = \frac{x-y}{2} > 0\). Since \(x_n \rightarrow x\), then there exists \(N_1 \in \mathbb{N}\) such that when \(n \geq N_1\)

$$ \begin{align*} |x_n - x| &< \epsilon \\ - \epsilon < x_n - x &< \epsilon \\ x - \epsilon < x_n &< x + \epsilon. \end{align*} $$

Hence

$$ \begin{align*} x_n > x - \epsilon = x - \frac{x-y}{2} = \frac{2x-(x-y)}{2} = \frac{x+y}{2}. \end{align*} $$

Similarly, since \(y_n \rightarrow y\), then there exists an \(N_2 \in \mathbb{N}\) such that when \(n \geq N_2\)

$$ \begin{align*} y_n < y + \epsilon = y + \frac{x-y}{2} = \frac{2y+(x-y)}{2} = \frac{x+y}{2} \end{align*} $$

Now, let \(N = \max\{N_0,N_1,N_2\}\). Assume \(n \geq N\). Then

$$ \begin{align*} y_n < \frac{x+y}{2} < x_n \end{align*} $$

This is a contradiction since for all \(n \geq N\), we know that \(x_n \leq y_n\). Therefore, we must have that \(x \leq y\). \(\ \blacksquare\)

References