More on Sequences — Question 1

Question 1 – Solution

Step 1: Identify the algorithm.

For $f(x)=x^2-3$, Newton’s method uses $x_{n+1}=x_n-f(x_n)/f'(x_n)$. Since $f'(x)=2x$, the update becomes $$x-\frac{f(x)}{f'(x)}=x-\frac{x^2-3}{2x}=\frac12\left(x+\frac3x\right),$$ which is exactly the given recurrence.

Step 2: Establish positivity and a lower bound.

All terms are positive because $a_1=2>0$ and, whenever $a_n>0$, both $a_n$ and $3/a_n$ are positive. More precisely, $$a_{n+1}-\sqrt3=\frac{(a_n-\sqrt3)^2}{2a_n}\ge0,$$ The right side is nonnegative, so $a_{n+1}\ge\sqrt3$. Since $a_1\ge\sqrt3$, induction proves $a_n\ge\sqrt3$ for every $n$.

Step 3: Prove monotonicity.

Using the lower bound, $$a_{n+1}-a_n=\frac{3-a_n^2}{2a_n}\le0,$$ Thus $a_{n+1}\le a_n$, so $(a_n)$ is decreasing.

Step 4: Prove convergence before solving for the limit.

The sequence is decreasing and bounded below by $\sqrt3$. The Monotone Convergence Theorem therefore guarantees $a_n\to L$ for some $L\ge\sqrt3$. Continuity and $L>0$ allow passage to the limit: $$L=\frac12\left(L+\frac3L\right),\qquad L^2=3.$$ The equation gives $L=\pm\sqrt3$, and the established positive lower bound selects $L=\sqrt3$.