Let $a_1 = 1$ and $a_{n+1} = \frac{a_n}2 + \frac2{a_n}$, $n ≥ 1$. Find the limit of the sequence $\left \{a_n \right \}$ if it exists.

Question

Should monotonic sequence theorem be used? Thanks in advance.

Answer 1

First, it's trivial by Mathematical Induction that $a_n \ge 0$.

Assuming the limit exists, you have

$$x=\frac{x}{2}+\frac{2}{x}$$ $$\implies x=\pm 2$$

Therefore, $$\lim_{n\to\infty}a_n=2$$

Answer 2

Note: For $x,$ real, positive:

$2/x +x/2 \ge 2\sqrt{(2/x)(x/2)} =2$. (AM-GM)

$a_{n+1} -a_n = -a_n/2 +2/a_n = \dfrac {-a_n^2 +4}{2a_n}.$

$a_n > 0$, $n \in \mathbb{Z+}$, and

$a_n \ge 2$ for $n \gt 1.$

Hence $a_{n+1} - a_n \le 0.$

$a_n$ is decreasing and has a lower bound $2$, implies convergent.

Can you find the limit?

Hint: L= L/2 + 2/L , why?

Answer 3

Consider the function

$F(x)=x^k-c$, where $k>1$ is a positive integer and $c>0$

$F$ is differentiable and $F'(x)=kx^{k-1}$ $\forall$ $x$ $\in$ $\mathbb{R}$

We can prove that the Newton-Raphson method converges to its positive soltution given a starting point $a_0>0$

Then, if we take the sequence defined by $a_1=1$ and $a_{n+1}=a_n-\frac{F(a_n)}{F'(a_n)}$ converges to $\sqrt[k]{c}$

Substituting $F$ and $F'$ with their polynomial expressions, we can obtain $a_{n+1}=a_{n}-\frac{a_{n}^k-c}{ka_{n}^{k-1}}=(1-\frac{1}{k})a_{n}+\frac{c}{ka_{n}^{k-1}}$

Now take $k=2$ and $c=4$ and we'll get our original sequence.

We can then conclude that the sequence is convergent to $\sqrt{4}=2$