试证 \(\{n(\sin(n+1) -\sin n)\}\) 发散

elim

题：试证 \(\{n(\sin(n+1) -\sin n)\}\) 发散。

elim

题：试证 \(\{n(\sin(n+1) -\sin n)\}\) 发散
证：取定整数\(d\,(\ge 2)\)使\(\,u(m)={\small\dfrac{m\pi+\frac{1}{2}}{2^d}}\in (0,1)\;(m\in\{0,1\}).\)
\(\qquad\)据(参见这里), 有正整数严格增列\(\{n_{m,k}\}\)使得
\(\qquad\,n_{m,k}\pi-\lfloor n_{m,k}\pi\rfloor = u(m) +o(k)\to u(m)\;(k\to\infty)\)
\(\qquad\,\overset{_\,}{2^d} n_{m,k}\pi-m\pi=2^d\lfloor n_{m,k}\rfloor+\frac{1}{2}+o(k)\)
\(\qquad\)令\(\,n_k^+{\small=}2^d\lfloor n_{0,k}\pi\rfloor,\;n_k^-{\small=}2^d\lfloor n_{1,k}\pi\rfloor,\;\)则\(\,\cos(n_k^{\pm}{\small+}\frac{1}{2})\overset{k\to\infty}{\to}\pm 1\).
\(\because\quad n(\sin(n+1)-\sin n)=2n\cos(n+\frac{1}{2})\sin\frac{1}{2},\)
\(\therefore\quad\displaystyle\lim_{k\to\infty}n_k^{\pm}(\sin(n_k^{\pm}+1)-\sin (n_k^{\pm}))=\pm\infty.\,\)所论序列发散.

luyuanhong

楼上 elim 的帖子很好！已收藏。