某个sin数列存在趋于0、±∞的子列

hbghlyj

这帖的第2问的回答中省略了一步:

Let$$b_n = n\sin\left(\frac{n^2+1}n\frac\pi2\right)$$As $n \to +\infty$ we can consider three subsequences of $b_n$:
\begin{aligned}
b_{n_k} &\to +\infty\\
b_{m_k} &\equiv 0\\
b_{l_k} &\to -\infty
\end{aligned}We are not interested in finding $n_k$, $m_k$ and $l_k$, but it is fairly easy to prove that they exist. Hint: recall that $n \to +\infty$ and use the fact that $\sin x \in [-1, 1]$.

如何证明$n_k$, $m_k$ 与 $l_k$存在呢

色k

\[b_n=n\sin\left(\frac{n\pi}2+\frac\pi{2n}\right),\]
则当 `k` 为整数且 `k\to+\infty` 时
\begin{align*}
b_{4k+1}&=(4k+1)\sin\left(\frac\pi2+\frac\pi{8k+2}\right)\to+\infty,\\
b_{4k-1}&=(4k-1)\sin\left(-\frac\pi2+\frac\pi{8k-2}\right)\to-\infty,\\
b_{4k}&=4k\sin\frac\pi{8k}\to\frac\pi2,\\
b_{4k+2}&=(4k+2)\sin\left(\pi+\frac\pi{8k+4}\right)\to-\frac\pi2,
\end{align*}
可见楼主所引用的内容里，正负无穷的对，而零的不对。

hbghlyj

色k 发表于 2022-12-5 10:07
\
则当 `k` 为整数且 `k\to+\infty` 时
\begin{align*}

谢谢. 所以原题答案应该是$$\liminf a_n = \cosh(\pi/2),\qquad\limsup a_n = +\infty$$