Prove that $\lim_{n \rightarrow +\infty} v_{n+1} = \sqrt{2}$?

Question

How to prove that the sequence defined by $v_0 = 2$ and the recurrence relation $v_{n+1} = \dfrac{v_n}{2}+\dfrac{1}{v_n}$ is convergent, of limit $\sqrt{2}$ ?

Answer 1

Just for the fun, for this particular problem, it is possible to work out a closed form for $v_n$.

Let $\displaystyle\;u_n = \frac{v_n - \sqrt{2}}{v_n + \sqrt{2}} \iff v_n = \sqrt{2}\left(\frac{1+u_n}{1-u_n}\right)$, we have

$$u_{n+1} = \frac{v_{n+1}-\sqrt{2}}{v_{n+1}+\sqrt{2}} = \frac{v_n^2 - 2\sqrt{2}v_n + 2}{v_n^2 + 2\sqrt{2}v_n + 2} = \frac{(v_n-\sqrt{2})^2}{(v_n+\sqrt{2})^2} = u_{n}^2 $$ From this, it is easy to see for $n \in \mathbb{N}$, we have $$u_n = u_0^{2^n} = \left(\frac{2-\sqrt{2}}{2+\sqrt{2}}\right)^{2^n} = (\sqrt{2}-1)^{2^{n+1}} \implies v_n = \sqrt{2}\left(\frac{1+(\sqrt{2}-1)^{2^{n+1}}}{1-(\sqrt{2}-1)^{2^{n+1}}}\right) $$ Since $|\sqrt{2}-1|<1$, what is inside the bracket in RHS goes to $1$ as $n \to \infty$.
As a result, $v_n \to \sqrt{2}$ as $n \to \infty$.

Answer 2

For $x>\sqrt {2}$, let $f (x)=x/2+1/x $. $$f'(x)=\frac {x^2-2}{2x^2} . $$

$f $ is increasing from $(\sqrt {2},+\infty) $ to $(\sqrt {2},+\infty) $.

the sequence $(v_n) $ is then monotonic.

$v_1=1,5 <v_0=2 \implies (v_n) $ decreasing.

on the hand two, by induction it is easy to prove that $v_n\ge\sqrt {2} $ using $v_{n+1}=f (v_n)\ge f (\sqrt {2})=\sqrt {2} $

$(v_n) $ is monotonic and bounded below , is convergent to $L $ with $$L=\frac {L}{2}+\frac {1}{L} $$ or $$L=\sqrt {2} $$

Answer 3

Rewrite $$v_{n+1} = \frac{v_n}{2}+\dfrac{1}{v_n}$$ as $$v_{n+1}=v_n-\frac{v_n}{2}+\frac{1}{v_n}=v_n-\frac{v_n^2-2}{2v_n}$$ and recognize that $v_{n+1}$ is the $(n+1)^\text{th}$ iterate of Newton method for solving $f(v)=v^2-2=0$

Answer 4

By induction $v_n>0$ for all $n \in \mathbb{N}$. Even-more, by the Am-GM inequality,

$$v_{n+1} \geq 2 \sqrt{\frac{v_n}{2}\frac{1}{v_n}}=\sqrt{2}$$

One can show that the sequence is monotonically decreasing:

$$v_{n+1}-v_{n}=\frac{v_n^2+2}{2v_n}-\frac{2v_n^2}{2v_n}$$

$$=\frac{2-v_n^2}{2v_n} \leq 0$$

As $v_n^2 \geq 2$. Hence the sequence is bounded below and monotonically decreasing, it must have a nonnegative limit $L$. As $\lim v_n=\lim v_{n+1}$, we conclude $L=\sqrt{2}$.

Answer 5

$$v_{n+1}-\sqrt2=\frac{(v_n-\sqrt2)^2}{2v_n}\geq0$$ and $v_1>\sqrt2$, which says $v_n>\sqrt2$ for all $n$.

Now, since $$v_{n+1}-v_n=\frac{1}{v_n}-\frac{v_n}{2}=\frac{(\sqrt2-v_n)(\sqrt2+v_n)}{2v_n}<0,$$ we obtain that $\{v_n\}$ is decreasing.

Id est, there is $\lim\limits_{n\rightarrow+\infty}v_n$ and let this limit equal to $a$.

Thus, $$a=\frac{a}{2}+\frac{1}{a},$$ which gives $a=\sqrt2$.

Done!