Approximation for $\pi$

I have not seen it before. Note that $\pi = \sqrt{a} + a$ where $a = (1+2\,\pi -\sqrt {1+4\,\pi })/2$, and what you're saying is that a rational approximation of $a$ is $9/5$. In fact, we have a continued fraction $$ a = 1 + \dfrac{1}{1 + \dfrac{1}{3+ \dfrac{1}{1+\dfrac{1}{1139 + \ldots}}}}$$ and $1+1/(1+1/(3+1/1)) = 9/5$. The fact that the first omitted element, $1139$, is so large makes this a very good approximation: the error in approximating $a$ by $9/5$ is only about $3.5 \times 10^{-5}$. Four elements later comes $7574$, so an even better approximation is $1+1/(1+1/(3+1/(1+1/(1139+1/(1+1/(15+1/1)))))) = 174530/96963$ with error about $1.4 \times 10^{-14}$.

EDIT: Perhaps even more remarkable are $$ \eqalign{\pi - \sqrt{1 + \dfrac{47}{35} \pi} &\approx \dfrac{6}{7}\cr \pi - \sqrt{\dfrac{3}{5} + \dfrac{5}{2} \pi } &\approx \dfrac{216}{923}\cr}$$

corresponding to the continued fractions

$$ \eqalign{\pi - \sqrt{1 + \dfrac{47}{35} \pi} &= \dfrac{1}{1+ \dfrac{1}{6 + \dfrac{1}{126402+ \ldots}}}\cr \pi - \sqrt{\dfrac{3}{5} + \dfrac{5}{2} \pi} &= \dfrac{1}{4+\dfrac{1}{3+\dfrac{1}{1+\dfrac{1}{1+\dfrac{1}{1+\dfrac{1}{19+\dfrac{1}{133286+\ldots}}}}}}}\cr}$$


Ramanujan found this approximation, among many others, according to Wolfram MathWorld equation 21 in linked page.