How to prove $\lim\limits_{x \to \infty} \frac{\pi\sqrt{x}}{\sin{(\frac{\pi}{\sqrt{x}} })} - x = \frac{\pi^2}{6}$?

Question

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?

Answer 1

$$y=\frac{\pi\sqrt{x}}{\sin \left(\frac{\pi }{\sqrt{x}}\right)} - x $$ $$x=\frac {\pi^2}{v^2} \quad \implies \quad y=\pi^2 \left(\frac {1}{v\sin(v) }-\frac {1}{v^2}\right)$$

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 $x$ $$y=\frac{\pi ^2}{6}+\frac{7 \pi ^4}{360 x}+O\left(\frac{1}{x^2}\right)$$

Answer 2

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

$$\frac{\pi\sqrt{x}}{\sin{(\frac{\pi}{\sqrt{x}} })} - x = \frac{\pi\sqrt{x}}{\frac{\pi}{\sqrt{x}}}\left(1+\frac16\frac{\pi^2}{x}+O\left(\frac{1}{x^2} \right)\right)-x=\frac{\pi^2}6+O\left(\frac{1}{x} \right)\to \frac{\pi^2}6$$

Answer 3

For simplicity let $x = \frac{1}{u^2}$. Then the limit becomes,

$$\lim_{u \to 0} (\frac{\pi u - \sin(\pi u)}{u^2 \sin(\pi u)})$$

This has indeterminate form of $\frac{0}{0}$, so we can apply l'hospitals rule,

$$=\lim_{u \to 0} (\frac{\pi - \pi \cos(\pi u)}{2u \sin(\pi u) + \pi u^2 \cos(\pi u)}) = \pi \lim_{u \to 0} (\frac{1 - \cos(\pi u)}{2u \sin(\pi u) + \pi u^2 \cos(\pi u)})$$

This has indeterminate form of $\frac{0}{0}$, so we can apply l'hospitals rule again,

$$=\pi \lim_{u \to 0} (\frac{\pi \sin(\pi u)}{2 \sin(\pi u) + 4 \pi u \cos(\pi u) - \pi^2u^2\sin(\pi u)}) = \pi^2 \lim_{u \to 0} (\frac{1}{2 + 4 \pi u \cot(\pi u) - \pi^2u^2})$$

Setting $v = \pi u$ simplifies the limit to,

$$=\pi^2 \lim_{v \to 0} (\frac{1}{2 + 4 v \cot(v) - v^2})$$

Now using the Maclaurin expansion of $\cot(v)$,

$$\pi^2 \lim_{v \to 0} (\frac{1}{2 + 4v(\frac{1}{v}-\frac{v}{3}-\frac{v^3}{45} - ...) - v^2}) = \pi^2 \lim_{v \to 0} (\frac{1}{2 + 4-\frac{4v^2}{3}-\frac{4v^4}{45} - ... - v^2})$$

Now setting $v=0$ the limit reduces to,

$$=\pi^2 (\frac{1}{2 + 4 - 0 - 0 - ... - 0}) = \frac{\pi^2}{6}$$

As required.