Final answer:
To prove the convergence of the sequence {a1=√6, an+1=√(6+an)}, we can show that it is bounded above by 3 and increasing. We find the limit of the sequence by solving a recursive equation and obtain a limit of 3. We also prove the upper bound using induction. Therefore, the sequence converges to a limit of 3.
Step-by-step explanation:
The given sequence is defined recursively as {a1=√6, an+1=√(6+an)} for all n ∈ N. To prove that the sequence converges, we need to show that it is bounded and monotonic. First, let's assume the sequence is increasing and find conditions for its upper bound. Taking the limit of the recursive equation, we get lim(n->∞) an+1 = lim(n->∞) √(6+an). Since the limit of the sequence equals the limit of its next term, we have lim(n->∞) an = lim(n->∞) √(6+an), which can be simplified to lim(n->∞) an = lim(n->∞) √(6+√(6+an-1)). Solving for the limit, we obtain lim(n->∞) an = √6 + √(6 + √(6 + ...)).
Let x = √6 + √(6 + √(6 + ...)). Then we can write the equation as x = √6 + √(6 + x). Squaring both sides and solving for x, we get x^2 - x - 6 = 0. Factoring the quadratic equation, we have (x-3)(x+2) = 0. Therefore, x can be either 3 or -2. Since the sequence begins with a positive value, we can disregard the negative solution. Hence, the limit is lim(n->∞) an = 3.
Now let's prove that the sequence is bounded above by 3. We can do this by induction. For the base case, a1 = √6 < 3. For the inductive step, assuming that an < 3, we will show that an+1 < 3. By squaring both sides of the recursive equation, we get an+1^2 < 6 + an. Since an is less than 3, it follows that an+1^2 < 9, and thus an+1 < 3. Therefore, the sequence is bounded above by 3, and by the Monotone Convergence Theorem, it must converge to a limit of 3.