Mostrar que $\,\,\, \lim_{n\to\infty} \sin\bigl(\pi\sqrt{n^2+1}\bigr)=0 $

Question

Alguien me puede ayudar a resolver este problema?

Mostrar que $$ \lim_{n\to\infty} \sin\bigl(\pi\sqrt{n^2+1}\bigr)=0 $$

Answer 1

Tenemos que $$ 0<\delta_n=\sqrt{n^2+1}-n=\frac{\big(\sqrt{n^2+1}+n\big)\big(\sqrt{n^2+1}-n\big)}{\sqrt{n^2+1}+n}= \frac{1}{\sqrt{n^2+1}+n}<\frac{1}{2n} $$ Así $$ \sin\left(\pi\sqrt{n^2+1}\right)=\sin\left(\pi\sqrt{n^2+1}-\pi n+\pi n\right)=\sin(\pi\delta_n+\pi n)=(-1)^n\sin(\pi\delta_n), $$ y como $\lvert \sin x\rvert\le \lvert x\rvert$, luego $$ \big|\sin\left(\pi\sqrt{n^2+1}\right)\big|=\lvert\sin(\pi\delta_n)\rvert\le \lvert\pi \delta_n\rvert<\frac{\pi}{2n}\a 0. $$ Nota. De hecho, uno puede usar lo anterior para mostrar que $$ \lim_{n\to\infty} (-1)^n\sin\big(\pi\sqrt{n^2+1}\big)=\frac{\pi}{2}. $$

Answer 2

Asumiendo $n$ a ser entero,

$$\sin\left(\pi\sqrt{n^2+1}\right)=(-1)^{n-1}\sin\left(n\pi-\pi\sqrt{n^2+1}\right)$$

$$=(-1)^{n-1}\sin\left[\pi(n-\sqrt{n^2+1})\right]$$

$$=(-1)^{n-1}\sin\left[\pi\left(\frac{n^2-(n^2+1)}{n+\sqrt{n^2+1}}\right)\right]$$

$$=(-1)^{n-1}\sin\left[-\pi\left(\frac1{n+\sqrt{n^2+1}}\right)\right]$$

Ahora $\displaystyle\lim_{n\to\infty}\left(\frac1{n+\sqrt{n^2+1}}\right)=0$

Answer 3

$$\lim_{n\to\infty}\sin\pi\sqrt{n^2+1}=\lim_{n\to\infty}(-1)^n\sin(\sqrt{n^2+1}-n)\pi$$ $$=\lim_{n\to\infty}(-1)^n\sin(\frac{1}{\sqrt{n^2+1}+n})\pi=\lim_{n\to\infty}(-1)^n\sin0\cdot\pi$$ $$=0$$