<s>[INST] How to prove $\lim\limits_{x \to \infty} \frac{\pi\sqrt{x}}{\sin{(\frac{\pi}{\sqrt{x}} })} - x = \frac{\pi^2}{6}$? As we know, the result of some expressions and series is equal to $\frac{\pi^2}{6} $ that the most important of them is $\zeta(2)$
Now, I have founded an equation whose limit at point $x=\infty$ is equal to $\frac{\pi^2}{6} : $
$$\lim_{x \to \infty}  \frac{\pi\sqrt{x}}{\sin{(\frac{\pi}{\sqrt{x}}  })} - x = \frac{\pi^2}{6}$$
I want to know how can we prove it?
 [/INST] We have that
$$\sin \left(\frac{\pi}{\sqrt{x}}\right)=\frac{\pi}{\sqrt{x}}-\frac16\frac{\pi^3}{x\sqrt{x}}+O\left(\frac{1}{x^2\sqrt{x}} \right)$$
and then
$$\frac{\pi\sqrt{x}}{\sin{(\frac{\pi}{\sqrt{x}} })} - x = \frac{\pi\sqrt{x}}{\frac{\pi}{\sqrt{x}}}\left(1+\frac16\frac{\pi^2}{x}+O\left(\frac{1}{x^2} \right)\right)-x=\frac{\pi^2}6+O\left(\frac{1}{x} \right)\to \frac{\pi^2}6$$
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