Compute limit if it exist of $a_1 = 2$, $a_{n+1} = \frac{1}{2} (a_n + \frac{1}{a_n})$

Question

$$a_1 = 2$$ $$a_{n+1} = \frac{1}{2} (a_n + \frac{1}{a_n})$$

I tried to use induction but it looks as if the sequence isn't monotonic.

Answer 1

Using AM-GM inequality we have $a_n \ge 1$ for all $n \ge 1 \implies 0 < \dfrac{1}{a_n} \le a_n \implies a_{n+1} \le a_n$. Thus $\{a_n\}$ is decreasing and bounded below by $1$, hence converges to $L$ that satisfies: $ L = \dfrac{L+\dfrac{1}{L}}{2} \implies L = 1$.

Answer 2

Hint: This sequence represents a common method of approximating a quantity. What method and what quantity is it?

Hint 2: A sequence $a_n = f^n(y)$, where $f^n$ means $f$ iterated $n$ times, often has its limit $x$ satisfy $x=f(x)$.

Answer 3

we have $a_1=2$

by induction $2\geq a_n>1$.

$$a_{n+1}=a_n-\frac{a_n^2-1}{2a_n}$$

$$a_{n+1}-1=(a_n-1)(1-\frac{a_n+1}{2a_n})$$

$$=\frac{1}{2}(a_n-1)(1-\frac{1}{a_n})$$

$\implies$

$$|a_{n+1}-1| \leq \frac{1}{2}|a_n-1|$$

$....\leq (\frac{1}{2})^n|a_1-1|$

this proves $(a_n)$ tends to $1$.

Answer 4

Let $\displaystyle\;b_n = \frac{a_n + 1}{a_n - 1}$, we have $$ b_{n+1} = \frac{a_{n+1}+1}{a_{n+1}-1} = \frac{a_n + \frac{1}{a_n} + 2}{a_n + \frac{1}{a_n} - 2} = \left(\frac{a_n + 1}{a_n - 1}\right)^2 = b_n^2$$

Together with $\displaystyle\;b_1 = \frac{a_1-1}{a_1+1} = \frac{2+1}{2-1} = 3$, we get

$$b_n = b_1^{2^{n-1}} = 3^{2^{n-1}} \quad\implies\quad a_n = \frac{b_n + 1}{b_n - 1} = \frac{3^{2^{n-1}} + 1}{3^{2^{n-1}} - 1} \to 1 \quad\text{ as }\quad n \to \infty $$

Answer 5

If you JUST want to find the limit the easiest thing to do is solve $$L=\frac{1}{2}\left(L+\frac{1}{L}\right)$$ you get $L=\pm 1$, and since $a_n$ is positive, $L=1$.

If you want to show the limit exists, it is decreasing. $$a_{n+1}=\frac{1}{2}\left(a_n+\frac{1}{a_n}\right)\leq a_n$$ is equivalent to $1\leq a_n^2$ and this is true because $$1\leq \frac{1}{2}\left(x+\frac{1}{x}\right)$$ for all $x$.