Solution: We will use induction to show that $s_n < 2$. Note that $s_1 = \sqrt{2} < 2$. Let for $s_k < 2$. We need to show that $s_{k+1} < 2$. Consider \begin{align*} s_{k+1} = \sqrt{2 + \sqrt{s_k} } \implies s_{k+1} ^2 = 2 + \sqrt{s_k} < 4. \end{align*} Thus, $s_{k+1} < 2$. Hence, by induction, we conclude that $s_n < 2$ for any $n \in \mathbb{N} $.

We now claim that $(s_n)$ is an increasing sequence. We will prove that $s_{n+1} - s_n \geq 0$. Again, we will use induction to prove our claim. For $n = 1$, \[ s_2 - s_1 = \sqrt{2 + \sqrt{2} } - \sqrt{2} > 0. \] Let $s_k - s_{k-1} \geq 0$. Now consider \begin{align*} s_{k+1} ^2 - s_k^2 & = \left(2 + \sqrt{s_k} \right) - \left( 2 + \sqrt{s_{k-1} } \right) \\ & = \sqrt{s_k} - \sqrt{s_{k-1}}. \end{align*} Since $s_n$ is positive for any $n \in \mathbb{N} $ \text{ and } hence $s_n \geq s_{n-1} $ implies $\sqrt{s_n} \geq \sqrt{s_{n-1} }.$ Thus, \begin{align*} s_{k+1} ^2 - s_k^2 \geq 0 \implies (s_{k+1} + s_k)(s_{k+1} - s_k) \geq 0. \end{align*} Since $s_{k+1} + s_k > 0$, we must have $s_{k+1} - s_k \geq 0$. Hence by the method of induction, we conclude $(s_n)$ is an increasing sequence.

Thus, we showed that $(s_n)$ is an increasing sequence which is bounded above and hence convergent.