Sequences — Question 8

Question 8 – Solution

Step 1: Prove an invariant bound.

We prove $0\le a_n<2$ by induction. The base case holds because $a_1=0$. Assume $0\le a_n<2$. Then $$0<2+a_n<4\quad\Longrightarrow\quad0<a_{n+1}=\sqrt{2+a_n}<2.$$ Thus the bound holds for $a_{n+1}$, completing the induction.

Step 2: Prove monotonicity.

Use induction again. Since $a_2=\sqrt2>a_1$, the base case holds. If $a_n>a_{n-1}$, the square-root function is strictly increasing, so $$a_{n+1}=\sqrt{2+a_n}>\sqrt{2+a_{n-1}}=a_n.$$ Thus $(a_n)$ is increasing. Together with the upper bound $2$, the Monotone Convergence Theorem guarantees $a_n\to L$ for some $0\le L\le2$.

Step 3: Solve the fixed-point equation.

Only after proving convergence may we pass to the limit in the recurrence. Continuity of the square root gives $$L=\sqrt{2+L}\quad\Longrightarrow\quad L^2-L-2=0 \quad\Longrightarrow\quad(L-2)(L+1)=0.$$ The algebraic candidates are $L=2$ and $L=-1$. Because every term and hence its limit are nonnegative, $L=-1$ is inadmissible. Therefore $L=2$. Solving this equation alone would identify only possible limits; boundedness and monotonicity are what prove a limit actually exists.