<s>[INST] A curious algebraic fraction that converges to $\frac{\sqrt{2}}{2}$ I have noticed that the  algebraic fraction 
$\frac{3a+2b}{4a+3b} $
Gives better and better approximations to $\sin  45^\circ = \frac{\sqrt{2}}{2} $
For $ a = b = 1$ we get $5/7 \approx  0.714 $
Now, taking $ a = 5, b = 7$, we get $ 29/41 \approx 0.707$
All the times, take 
$ a_{n+1} = 3a_n + 2b_n$
$ b_{n+1} = 4a_n+3b_n$ 
And this process converges to $\frac{\sqrt{2}}{2} $
I have arrived at this result by a casual consideration. When I take the approximation $5/7$, I wonder what happens if I rewrite the numbers $5$ and $7$ as a function of two quantities $a, b$  in order to get a better approximation by successive iterations, in which the first approximation is the case $ a = b = 1$. Thus I have written $ 5 = 3a + 2b$ and the number $7 = 4a+3b$ I have selected both numerator and denominator to be consecutive numbers, and surprisingly this gives better approximations to $\frac{\sqrt{2}}{2}$
The question now is: how to prove that, in fact, this  algebraic fraction, whit the described iteration above, converges to the desired value?
 [/INST] I think we can split the ratio $(3a + 2b)/(4a + 3b)$ further. Let $a/b$ be a rational approximation to $1/\sqrt{2}$. Then I prove that $(a + b)/(2a + b)$ is a better approximation to $1/\sqrt{2}$ but in opposite direction.
Clearly we have $$\left(\frac{a + b}{2a + b}\right)^{2} - \frac{1}{2} = \frac{b^{2} - 2a^{2}}{2(2a + b)^{2}} = \frac{b^{2}}{(2a + b)^{2}}\left(\frac{1}{2} - \frac{a^{2}}{b^{2}}\right)$$ so that $Y = \left(\dfrac{a + b}{2a + b}\right)^{2} - \dfrac{1}{2}$ and $X = \dfrac{a^{2}}{b^{2}} - \dfrac{1}{2}$ have opposite signs and $|Y| < |X|$ and thus the claim in previous paragraph is established.
Applying the transformation $a/b \to (a + b)/(2a + b)$ twice on $a/b$ we get $$\frac{3a + 2b}{4a + 3b}$$ and we are thus ensured that if $a/b$ is a rational approximation to $1/\sqrt{2}$ then $\dfrac{3a + 2b}{4a + 3b}$ is a better approximation to $1/\sqrt{2}$ in the same direction. Since a bounded and monotone sequence is convergent it follows that repeated application of the transformation $a/b \to (3a + 2b)/(4a + 3b)$ to any positive rational number $a/b$ converges to $1/\sqrt{2}$.

BTW this comes from an exercise in Hardy's A Course of Pure Mathematics (page 11, Examples III, problem 3) but Hardy uses it for $\sqrt{2}$:
Show that if $m/n$ is a good approximation to $\sqrt{2}$, then $(m + 2n)/(m + n)$ is a better one, and that the errors in the two cases are in opposite directions.
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