Properties of Convergent Sequences - The Squareroot Law

# Properties of Convergent Sequences - The Squareroot Law for Nonnegative Sequences

 Theorem 1: Let $(a_n)$ be a nonnegative sequence of real numbers (that is, $a_n \geq 0$ for all $n \in \mathbb{N}$. If $(a_n)$ converges to $a$ then $(\sqrt{a_n})$ converges to $\sqrt{a}$.
• Proof: Let $\epsilon > 0$ be given.
• Case 1: Suppose that $a = 0$. Since $(a_n)$ converges to $a = 0$ we have that for all $\epsilon > 0$ there exists an $N \in \mathbb{N}$ such that if $n \geq N$ then $|a_n - 0| < \epsilon^2$. But then $a_n = |a_n| < \epsilon^2$. Taking the squareroot of both sides gives us that for all $n \geq N$:
(1)
\begin{align} \quad \sqrt{a_n} = |\sqrt{a_n} - 0| < \epsilon \end{align}
• So $(\sqrt{a_n})$ converges to $0 = \sqrt{0} = \sqrt{a}$.
• Case 2: Suppose that $a > 0$. Since $(a_n)$ converges to $a > 0$ we have that for $\epsilon_1 = \sqrt{a} \epsilon > 0$ there exists an $N_1 \in \mathbb{N}$ such that if $n \geq N_1$ then:
(2)
\begin{align} \quad |a_n - a| < \epsilon_1 = \sqrt{a} \epsilon \quad (*) \end{align}
• Then for $N = N_1$ we have that for all $n \geq N$ that $(*)$ holds and:
(3)
\begin{align} \quad | \sqrt{a_n} - \sqrt{a} | = \frac{|a_n - a|}{|\sqrt{a_n} + \sqrt{a}|} \leq \frac{|a_n - a|}{\sqrt{a}} < \frac{\epsilon_1}{\sqrt{a}} = \frac{\sqrt{a} \epsilon}{\sqrt{a}} = \epsilon \end{align}
• So $(\sqrt{a_n})$ converges to $(\sqrt{a})$. $\blacksquare$