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?

$$y=\frac{\pi\sqrt{x}}{\sin \left(\frac{\pi }{\sqrt{x}}\right)} - x $$
$$x=\frac {\pi^2}{v^2} \quad \implies \quad y=\pi^2 \left(\frac {1}{v\sin(v) }-\frac {1}{v^2}\right)$$
By Taylor
$$\sin(v)=v-\frac{v^3}{6}+\frac{v^5}{120}+O\left(v^7\right)$$
$$v\sin(v)=v^2-\frac{v^4}{6}+\frac{v^6}{120}+O\left(v^8\right)$$ Long division
$$\frac {1}{v\sin(v) }=\frac{1}{v^2}+\frac{1}{6}+\frac{7 v^2}{360}+O\left(v^4\right) \quad \implies \quad y=\pi^2 \left(\frac{1}{6}+\frac{7 v^2}{360}+O\left(v^4\right)\right)$$ Back to $x$
$$y=\frac{\pi ^2}{6}+\frac{7 \pi ^4}{360 x}+O\left(\frac{1}{x^2}\right)$$
