一道数分题

请证明：级数 \(\sum_{n=1}^{\infty} \sin\frac{1}{n^p}\) 仅当 \(p>1\) 时收敛。

题解

引理：若 \(f\left(x\right)\sim x^p\)，则 \(f\left(x+1\right)-f\left(x\right)\sim x^{p-1}\)

证：由牛顿二项式定理，当 \(x\to \infty\) 时 \(\left(1+\frac{1}{x}\right)^{p-1}=\sum_{k=0}^{\infty}\binom{p-1}{k}1^{p-1-k}\left(\frac{1}{x}\right)^k=1+\left(p-1\right)\frac{1}{x}+\sum_{k=2}^{\infty}\binom{p-1}{k}\left(\frac{1}{x}\right)^k=1+\left(p-1\right)\frac{1}{x}+o\left(\frac{1}{x}\right)\)

\(\therefore\left(x+1\right)\left(1+\frac{1}{x}\right)^{p-1}=x+1+p-1+\frac{p-1}{x}+\left(x+1\right)o\left(\frac{1}{x}\right)=x+p+o\left(1\right)\)

\(\therefore\left(x+1\right)\left(1+\frac{1}{x}\right)^{p-1}-x\sim p\)，故 \(\left(x+1\right)^p-x^p\sim px^{p-1}\)

\(\because f\left(x+1\right)\sim \left(x+1\right)^p,f\left(x\right)\sim x^p\)

\(\therefore f\left(x+1\right)-f\left(x\right)\sim \left(x+1\right)^p-x^p\sim px^{p-1}\sim x^{p-1}\)，引理得证

回到原题

\(1^\circ\) \(p>1\) 时，\(0<\frac{1}{n^p}\le 1\)，故 \(\sin\frac{1}{n^p}<\frac{1}{n^p}\)

\(\therefore\sum_{n=1}^{\infin}\sin\frac{1}{n^p}<\sum_{n=1}^{\infin}\frac{1}{n^p}\)

\(p>1\) 时 \(\sum_{n=1}^{\infin}\frac{1}{n^p}\) 收敛，故 \(\sum_{n=1}^{\infin}\sin\frac{1}{n^p}\) 收敛

\(2^\circ\) \(0\le p<1\) 时，\(0<\frac{1}{n^p}\le 1\)

由 \(x\in\left(0,1\right]\) 时 \(\sin x>x-\frac{x^3}{3!}=x-\frac{x^3}{6}\) 知 \(\sin\frac{1}{n^p}>\frac{1}{n^p}-\frac{1}{6n^{3p}}=\frac{1}{n^p}\left(1-\frac{1}{6n^{2p}}\right)\)

\(\because \frac{1}{n^{2p}}\le 1\)，故 \(1-\frac{1}{6n^{2p}}\ge \frac{5}{6}\)，即 \(\sin\frac{1}{n^p}>\frac{5}{6n^p}\)

\(\therefore\sum_{n=1}^{\infin}\sin\frac{1}{n^p}>\sum_{n=1}^{\infin}\frac{5}{6n^p}=\frac{5}{6}\sum_{n=1}^{\infin}\frac{1}{n^p}\)

\(0\le p<1\) 时 \(\sum_{n=1}^{\infin}\frac{1}{n^p}\) 发散，故 \(\sum_{n=1}^{\infin}\sin\frac{1}{n^p}\) 发散

\(3^\circ\) \(p<0\) 时，要证 \(\sum_{n=1}^{\infin}\sin\frac{1}{n^p}\) 发散，等价于证 \(p>0\) 时，\(\sum_{n=1}^{\infin}\sin n^p\) 发散

反证：若 \(\sum_{n=1}^{\infin}\sin n^p\) 发散不成立，则 \(\sum_{n=1}^{\infin}\sin n^p\) 收敛，故 \(\lim\limits_{N\to\infty}\sum_{n=1}^{N}\sin n^p\) 存在

由柯西收敛准则知，\(\forall \epsilon > 0\)，\(\exists N\in \mathbb N\) 使 \(\forall m,k> N\) 有 $\left|\sum_{n=1}^{m}\sin n^p -\sum_{n=1}^{k}\sin n^p\right|<\epsilon $

取 \(m=k+1\)，即 \(\forall k>N\) 有 \(\left|\sin\left(k+1\right)^p\right|<\epsilon\)

\(\therefore \forall \epsilon>0,\exists N\in \mathbb{N}\) 使 \(\forall n\ge N\) 有 \(\left|\sin n^p\right|<\epsilon\)

对于 \(\forall x\in \mathbb R\)，\(\exists\) 唯一 \(k,r\) 使得 \(k\in\mathbb{Z},r\in\left(-\frac{\pi}{2},\frac{\pi}{2}\right]\) 且 \(x=k\pi+r\)，记 \(k\left(x\right)=k,r\left(x\right)=r\)

\(\therefore \forall \epsilon>0,\exists N\in \mathbb{N}\) 使 \(\forall n\ge \mathbb{N}\) 有 \(\left|\sin n^p\right|=\left|\sin r\left(n^p\right)\right|<\epsilon\)，即 \(\lim\limits_{n\to\infty}\sin r\left(n^p\right)=0\)

又 \(\because -\frac{\pi}{2}<r\left(n^p\right)\le \frac{\pi}{2}\) 且 \(\sin x\) 在 \(\left(-\frac{\pi}{2},\frac{\pi}{2}\right]\) 上单增，\(\sin 0=0\)

\(\therefore \lim\limits_{n\to\infty}r\left(n^p\right)=0\)

\(\therefore \forall \epsilon>0,\exists N\in\mathbb{N}\) 使 \(\forall n>N\) 有 \(\left|r\left(n^p\right)\right|<\epsilon\)

\(\therefore \forall \epsilon>0\) 且 \(\epsilon<\frac{\pi}{2}\)，令 \(\epsilon_1=\frac{\epsilon}{2^{\left[p\right]+1}}\)，\(\exists N_1\in \mathbb{N}\) 使 \(\forall n\ge \mathbb{N}\) 有 \(\left|r\left(n^p\right)\right|<\epsilon_1\) \(\left(*\right)\)

令 \(f_p\left(x\right)=x^p\)，\(f_{p-1}\left(x\right)=f_p\left(x+1\right)-f_p\left(x\right)\)，\(f_{p-2}\left(x\right)=f_{p-1}\left(x+1\right)-f_{p-1}\left(x\right)\)，以此类推

\(\forall n\ge N_1,f_p\left(n\right)=n^p=k\left(n^p\right)\pi+r\left(n^p\right)\)，且 \(\left|r\left(n^p\right)\right|<\epsilon_1\)，即 \(\left|r\left(f_p\left(n\right)\right)\right|<\epsilon_1\)

\(p\ge 0\) 时：\(f_{p-1}\left(n\right)=f_p\left(n+1\right)-f_p\left(n\right)=\left(k\left(f_{p}\left(n+1\right)\right)-k\left(f_{p}\left(n\right)\right)\right)\pi+\left(r\left(f_{p}\left(n+1\right)\right)-r\left(f_{p}\left(n\right)\right)\right)\) 且 \(\left|r\left(f_{p}\left(n+1\right)\right)-r\left(f_{p}\left(n\right)\right)\right|\le\left|r\left(f_{p}\left(n+1\right)\right)\right|+\left|r\left(f_{p}\left(n\right)\right)\right|<2\epsilon_1<\epsilon<\frac{\pi}{2}\)，故 \(r\left(f_{p-1}\left(n\right)\right)=r\left(f_{p}\left(n+1\right)\right)-r\left(f_{p}\left(n\right)\right),\left|r\left(f_{p-1}\left(n\right)\right)\right|<2\epsilon_1\)

\(p\ge 1\) 时：\(f_{p-2}\left(n\right)=f_{p-1}\left(n+1\right)-f_{p-1}\left(n\right)=\left(k\left(f_{p-1}\left(n+1\right)\right)-k\left(f_{p-1}\left(n\right)\right)\right)\pi+\left(r\left(f_{p-1}\left(n+1\right)\right)-r\left(f_{p-1}\left(n\right)\right)\right)\) 且 \(\left|r\left(f_{p-1}\left(n+1\right)\right)-r\left(f_{p-1}\left(n\right)\right)\right|\le\left|r\left(f_{p-1}\left(n+1\right)\right)\right|+\left|r\left(f_{p-1}\left(n\right)\right)\right|<4\epsilon_1<\epsilon<\frac{\pi}{2}\)，故 \(r\left(f_{p-2}\left(n\right)\right)=r\left(f_{p-1}\left(n+1\right)\right)-r\left(f_{p-1}\left(n\right)\right),\left|r\left(f_{p-2}\left(n\right)\right)\right|<4\epsilon_1\)

以此类推，由 \(p\ge \left[p\right]\gt \left[p\right]-1\) 可知

由 \(p\ge \left[p\right]\)，有 \(\left|r\left(f_{p-\left[p\right]-1}\left(n\right)\right)\right|<2^{\left[p\right]+1}\epsilon_1=\epsilon\) \(\left(1\right)\)

由 \(p\ge \left[p\right]-1\)，有 \(\left|r\left(f_{p-\left[p\right]}\left(n\right)\right)\right|<2^{\left[p\right]}\epsilon_1=\frac{\epsilon}{2}\) \(\left(2\right)\)

结合引理知 \(f_p\left(n\right)\sim n^p,f_{p-1}\left(n\right)\sim n^{p-1},f_{p-2}\left(n\right)\sim n^{p-2},\cdots,f_{p-\left[p\right]}\left(n\right)\sim n^{p-\left[p\right]},f_{p-\left[p\right]-1}\left(n\right)\sim n^{p-\left[p\right]-1}\)

当 \(p\in\mathbb{Z}\) 时，\(p-\left[p\right]=0\)，\(f_p\left(x\right)\) 中最高项系数为 \(1\)，\(f_{p-1}\left(x\right)\) 中最高项系数为 \(p\)，\(f_{p-2}\left(x\right)\) 中最高项系数为 \(p\left(p-1\right)\)，…，\(f_{0}\left(x\right)\) 中最高项系数为 \(p!\)，而 \(f_0\left(x\right)\) 只有常数项，故 \(f_0\left(x\right)=p!\) 恒成立

\(\because f_0\left(x\right)\in \mathbb{Z^+}\)

\(\therefore \left|r\left(f_{p-\left[p\right]}\left(n\right)\right)\right|=\left|r\left(p!\right)\right|>0\)，取 \(\epsilon=\left|r\left(p!\right)\right|\) 即有 \(\left(2\right)\) 式不成立，矛盾！

当 \(p\not\in\mathbb{Z}\) 时

\(\because p-\left[p\right]-1\in\left(-1,0\right)\)

\(\therefore n\to\infty\) 时，\(n^{p-\left[p\right]-1}\to0,f_{p-\left[p\right]-1}\left(n\right)\to0\)，即 \(n\to\infty\) 时 \(k\left(f_{p-\left[p\right]-1}\left(n\right)\right)=0\) 且 \(r\left(f_{p-\left[p\right]-1}\left(n\right)\right)\to0\)

\(\therefore \forall \epsilon_2>0\)，\(\exists N_2\in\mathbb{N}\) 使 \(\forall n\ge N_2\) 有 \(\left|f_{p-\left[p\right]-1}\left(n\right)\right|<\epsilon_2\) \(\left(**\right)\)

对 \(\forall\epsilon\in\left(0,\frac{\pi}{2}\right)\)，令 \(\epsilon_2=\frac{\epsilon}{2}\)，则 \(\exists N_2\in \mathbb{N}\) 满足 \(\left(**\right)\)，当 \(\epsilon_1=\frac{\epsilon}{2^{\left[p\right]+1}}\) 时，\(\exists N_1\in \mathbb{N}\) 满足 \(\left(*\right)\)

令 \(N=\max\left(N_2,N_1\right)\)，则 \(N\) 同时满足 \(\left(*\right)\) 和 \(\left(**\right)\)

\(\forall n\ge \mathbb{N}\)，\(f_{p-\left[p\right]}\left(n+1\right)=f_{p-\left[p\right]}\left(n\right)+f_{p-\left[p\right]-1}\left(n\right)=k\left(f_{p-\left[p\right]}\left(n\right)\right)\pi+\left(r\left(f_{p-\left[p\right]}\left(n\right)\right)+r\left(f_{p-\left[p\right]-1}\left(n\right)\right)\right)\)

由于 \(\left|r\left(f_{p-\left[p\right]}\left(n\right)\right)+r\left(f_{p-\left[p\right]-1}\left(n\right)\right)\right|\le\left|r\left(f_{p-\left[p\right]}\left(n\right)\right)\right|+\left|r\left(f_{p-\left[p\right]-1}\left(n\right)\right)\right|<2^{\left[p\right]}\epsilon_1+\epsilon_2=\frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon<\frac{\pi}{2}\)

\(\therefore r\left(f_{p-\left[p\right]}\left(n+1\right)\right)=r\left(f_{p-\left[p\right]}\left(n\right)\right)+r\left(f_{p-\left[p\right]-1}\left(n\right)\right)\)，故 \(k\left(f_{p-\left[p\right]}\left(n+1\right)\right)=k\left(f_{p-\left[p\right]}\left(n\right)\right)\)

\(\therefore\forall n\ge {N}\) 有 \(k\left(f_{p-\left[p\right]}\left(n\right)\right)=k\left(f_{p-\left[p\right]}\left(N\right)\right)\)，故 \(f_{p-\left[p\right]}\left(n\right)=k\left(f_{p-\left[p\right]}\left(N\right)\right)\pi+r\left(f_{p-\left[p\right]}\left(n\right)\right)\in\left(k\left(f_{p-\left[p\right]}\left(N\right)\right)\pi-\frac{\pi}{2},k\left(f_{p-\left[p\right]}\left(N\right)\right)\pi+\frac{\pi}{2}\right]\)

\(\therefore\lim\limits_{n\to\infty}f_{p-\left[p\right]}\left(n\right)\in\left(k\left(f_{p-\left[p\right]}\left(N\right)\right)\pi-\frac{\pi}{2},k\left(f_{p-\left[p\right]}\left(N\right)\right)\pi+\frac{\pi}{2}\right]\) \(\left(3\right)\)

另一方面，\(f_{p-\left[p\right]}(n)\sim n^{p-\left[p\right]}\)，由 \(p-\left[p\right]\in\left(0,1\right)\) 知 \(n\to\infty\) 时 \(n^{p-\left[p\right]}\to \infty\)，故 \(\lim\limits_{n\to\infty}f_{p-\left[p\right]}\left(n\right)=\infty\) \(\left(4\right)\)

\(\left(3\right)\left(4\right)\) 矛盾！

\(\therefore p<0\) 时，\(\sum_{n=1}^{\infin}\sin\frac{1}{n^p}\) 发散

综上所述， \(\sum_{n=1}^{\infin}\sin\frac{1}{n^p}\) 收敛当且仅当 \(p>1\)