An increasing rational sequence converging to $\sqrt{2} $ Can you suggest an increasing rational sequence converging to $\sqrt{2} $ in its closed form?
Can you make it from the recursion $x_{n+1}=\frac{x_n}{2}+\frac{1}{x_n}$? I dont think so, because, in that case $x_n$ should be $\sqrt{2} $ enough (Not interested). Further, how can we write the sequence $1,1.4,1.41,1.414,...$ in general form?
 A: You are right, the sequence $a_0=1,a_1=1.4,a_2=1.41,a_3=1.414,...$ does increase to $\sqrt2$. The $n$-th term of the sequence, specifically, is $\sqrt2$ rounded down to $n$ decimal places. Note that, to round down, we use the floor function:
$$\lfloor 1.9\rfloor=1.$$
Also note that if we multiply $\sqrt2$ by $10^n$ and round down we get an integer with $n$ digits which is less than $10^n\sqrt2$. Now, all we do is multiply our remaining integer by $10^{-n}$ to put the decimal point back were it's supposed to be. In symbols,
$$a_n=10^{-n}\lfloor10^n\sqrt2\rfloor.$$
Note that for all $n$, $$1\le a_n<\sqrt2$$
and $a_n\le a_{n+1}$, so $\lim_{n\to\infty}a_n=\sqrt{2}$.

Yes, there are more efficient and correct methods for the calculation of square roots, but this is exactly what was asked for: an increasing rational sequence converging to $\sqrt2$.
A: If we take $a_1=2$, and $a_{n+1}=\frac12(a_n+2/a_n)$ we get a famous decreasing
sequence of rationals converging to $\sqrt2$. So instead take $b_n=2/a_n$.
Now the $b_n$ increase and tend to $\sqrt2$. In fact $b_1=1$ and
$$b_{n+1}=\frac{4}{b_n+2/b_n}.$$
$$ $$ $$ $$
A: Consider $f(x)=(x^2-2)(x-2)$. This has a root at $\sqrt{2}$, and $f'(\sqrt{2})<0$, and $f''(\sqrt{2})>0$. Therefore if you use Newton's root approximation method starting at a rational number just below $\sqrt{2}$, you will get a sequence of rational numbers converging to $\sqrt{2}$ from below.
The recursion is $x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}=x_n-\frac{(x_n^2-2)(x_n-2)}{3x_n^2-4x_n-2}$.
For example, $x_1=1$.
Then $x_2=x_1-\frac{f(x_1)}{f'(x_1)}=1-\frac{1}{-3}=\frac{4}{3}$.
Then $x_3=x_2-\frac{f(x_2)}{f'(x_2)}=\frac{4}{3}-\frac{4/27}{-2}=\frac{38}{27}$.
And $x_4=\cdots=\frac{23482}{16605}=1.41415236\ldots$ which already rounds to the same value as $\sqrt{2}$ at the ten-thousandths place.
And so on.
A: Why you think $x_{n+1}=\frac{1}{2}\left(x_n+\frac{2}{x_n}\right)$ will not work$?$ Hints: Show that $x_n^2\geq2$ (ignoring the initial term), and then use this to prove that $x_n-x_{n+1}\geq0$. Conclude that $\lim x_n=\sqrt{2}$
