1. **State the problem:** We want to prove by induction that the infinite nested radical expression $$x = \sqrt{2 + \sqrt{2 + \sqrt{2 + \cdots}}}$$ satisfies $$x < 2$$. 2. **Define the sequence:** Let $$a_1 = \sqrt{2}$$ and $$a_{n+1} = \sqrt{2 + a_n}$$ for $$n \geq 1$$. We want to show $$a_n < 2$$ for all $$n$$, and then conclude about the limit. 3. **Base case:** For $$n=1$$, $$a_1 = \sqrt{2} \approx 1.414 < 2$$. So the base case holds. 4. **Inductive hypothesis:** Assume $$a_k < 2$$ for some $$k \geq 1$$. 5. **Inductive step:** Show $$a_{k+1} < 2$$. We have $$ a_{k+1} = \sqrt{2 + a_k}. $$ Using the inductive hypothesis $$a_k < 2$$, $$ a_{k+1} < \sqrt{2 + 2} = \sqrt{4} = 2. $$ 6. **Conclusion:** By induction, $$a_n < 2$$ for all $$n$$. 7. **Limit:** Since $$a_n$$ is increasing and bounded above by 2, the limit $$x = \lim_{n \to \infty} a_n$$ exists and satisfies $$ x = \sqrt{2 + x}. $$ Squaring both sides, $$ x^2 = 2 + x. $$ Rearranged, $$ x^2 - x - 2 = 0. $$ Solving the quadratic, $$ x = \frac{1 \pm \sqrt{1 + 8}}{2} = \frac{1 \pm 3}{2}. $$ Since $$x > 0$$, we take $$ x = 2. $$ Thus, the infinite nested radical converges to 2, but each finite approximation $$a_n < 2$$. **Final answer:** The nested radical expression is always less than 2 for any finite number of radicals, and converges to 2 in the limit.