习题6.2

第 2 题

设 \(S(x) = \sum\limits_{n = 1}^{\infty}\dfrac{1}{2^n}\tan\dfrac{x}{2^n}\), 计算 \(\int_{\frac{\pi}{6}}^{\frac{\pi}{3}}S(x)\text{d}{x}\).

解

记 \(u_n(x) = \dfrac{1}{2^n}, v_n(x) = \tan\dfrac{x}{2^n}, I = \left(\dfrac{\pi}{6}, \dfrac{\pi}{3}\right)\). 由于 \(v_n(x)\) 对任意固定的 \(x\in I\) 均单调递减, 且 \(\left|v_n(x)\right| \le \dfrac{\sqrt{3}}{3}\) 在 \(I\) 上一致有界, 且 \(\sum\limits_{n = 1}^{\infty}u_n(x)\) 与 \(x\) 无关, 在 \(I\) 上一致收敛, 故由 Abel 判别法知 \(S(x)\) 在 \(I\) 上一致收敛. 故可以将积分与求和符号交换次序:

\[ \begin{aligned} \int_{\frac{\pi}{6}}^{\frac{\pi}{3}}S(x)\text{d}{x} & = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}}\sum\limits_{n = 1}^{\infty}\dfrac{1}{2^n}\tan\dfrac{x}{2^n}\text{d}{x} \\ & = \sum\limits_{n = 1}^{\infty}\int_{\frac{\pi}{6}}^{\frac{\pi}{3}}\dfrac{1}{2^n}\tan\dfrac{x}{2^n}\text{d}{x} \\ & = \sum\limits_{n = 1}^{\infty}\left(-\ln\cos\dfrac{x}{2^n}\right)\bigg\vert_{x=\frac{\pi}{6}}^{x=\frac{\pi}{3}} \\ & = \sum\limits_{n = 1}^{\infty}\ln\dfrac{\cos\frac{\pi}{3\cdot2^{n+1}}}{\cos\frac{\pi}{3\cdot2^n}} \\ & = \lim\limits_{n \to \infty}\ln\dfrac{\cos\frac{\pi}{3\cdot2^{n+1}}}{\cos\frac{\pi}{6}} \\ & = \ln 2 - \dfrac{\ln 3}{2} \end{aligned} \]

第 3 题

证明:

\[ \int_{0}^{1}x^x\text{d}{x} = 1 - \dfrac{1}{2^2} + \dfrac{1}{3^3} - \dfrac{1}{4^4} + \cdots + (-1)^n\dfrac{1}{(n + 1)^{n + 1}} + \cdots. \]

证明

由于 \(x^x = e^{x\ln x} = 1 + (x\ln x) + \dfrac{1}{2}(x\ln x)^2 + \cdots = \sum\limits_{n = 0}^{\infty}\dfrac{(x \ln x)^n}{n!}\), 且 \(\left|x \ln x\right| \in (0, 1), \forall x \in (0, 1)\), 故 \(\sum\limits_{n = 0}^{\infty}\dfrac{(x \ln x)^n}{n!} < \sum\limits_{n = 0}^{\infty}\dfrac{1}{n!}\). 而 \(\sum\limits_{n = 0}^{\infty}\dfrac{1}{n!}\) 收敛, 故由 Weierstrass 判别法知 \(\sum\limits_{n = 0}^{\infty}\dfrac{(x \ln x)^n}{n!}\) 在 \((0, 1)\) 上一致收敛. 因此可以交换积分与求和的次序:

\[ \begin{aligned} \int_0^1x^x\text{d}{x} & = \int_{0}^{1}\sum\limits_{n = 0}^{\infty}\dfrac{(x \ln x)^n}{n!}\text{d}{x} \\ & = \sum\limits_{n = 0}^{\infty}\int_{0}^{1}\dfrac{(x \ln x)^n}{n!}\text{d}{x} \\ & = \sum\limits_{n = 0}^{\infty}\dfrac{1}{n!}\int_{0}^{1}(x\ln x)^n\text{d}{x} \\ & = \sum\limits_{n = 0}^{\infty}\dfrac{1}{n!}\int_{0}^{1}\dfrac{\ln^n x}{n + 1}\text{d}{(x^{n + 1})} \\ & = \sum\limits_{n = 0}^{\infty}\left(\dfrac{x^{n + 1}\ln^n x}{n + 1}\bigg\vert_{0}^{1} - \int_{0}^{1}\dfrac{n x^n \ln^{n-1} x}{n + 1}\text{d}{x}\right) \\ & = -\sum\limits_{n = 0}^{\infty}\int_{0}^{1}\dfrac{n x^n \ln^{n-1} x}{n + 1}\text{d}{x} \\ & = \sum\limits_{n = 0}^{\infty}\int_{0}^{1}\dfrac{n(n-1) x^n \ln^{n-2} x}{(n + 1)^2}\text{d}{x} \\ & = \cdots \\ & = \sum\limits_{n = 0}^{\infty}(-1)^{n}\int_{0}^{1}\dfrac{n! x^n}{(n + 1)^n}\text{d}{x} \\ & = \sum\limits_{n = 0}^{\infty}\dfrac{(-1)^n n!}{(n + 1)^{n + 1}} \\ & = 1 - \dfrac{1}{2^2} + \dfrac{1}{3^3} - \dfrac{1}{4^4} + \cdots + (-1)^n\dfrac{1}{(n + 1)^{n + 1}} + \cdots \end{aligned} \]

命题得证.

第 4 题

证明: 函数 \(f(x) = \sum\limits_{n = 1}^{\infty}\dfrac{n}{x^n}\) 是 \((1, +\infty)\) 上的连续函数.

证明

要证明函数 \(f(x) = \sum\limits_{n = 1}^{\infty}\dfrac{n}{x^n}\) 是 \((1, +\infty)\) 上的连续函数, 只需证 \(f(x)\) 在任意 \(x = x_0 > 1\) 处连续即可. \(\forall x_0 > 1\), 取 \(r \in (1, r_0)\), 则 \(f(x) = \sum\limits_{n = 1}^{\infty}\dfrac{n}{x^n} \le \sum\limits_{n = 1}^{\infty}\dfrac{n}{r^n}\), 而 \(\sum\limits_{n = 1}^{\infty}\dfrac{n}{r^n}\) 收敛. 故由 Weierstrass 判别法知级数在 \([r, +\infty)\) 上一致收敛. 因此 \(f(x)\) 在任意 \(x = x_0 > 1\) 处连续. 命题得证.

第 5 题

证明: 级数 \(\sum\limits_{n = 1}^{\infty}\dfrac{x^2}{(1 + x^2)^n}\) 对任意的 \(x\) 绝对收敛, 但在 \((-\infty, +\infty)\) 上非一致收敛.

证明

由于 \(\sum\limits_{n = 1}^{\infty}\dfrac{x^2}{(1 + x^2)^n} < \sum\limits_{n = 1}^{\infty}\dfrac{1 + x^2}{(1 + x^2)^n} = \sum\limits_{n = 1}^{\infty}\dfrac{1}{(1 + x^2)^{n - 1}}\), 且 \(\sum\limits_{n = 1}^{\infty}\dfrac{1}{(1 + x^2)^{n - 1}}\) 绝对收敛, 因此 级数 \(\sum\limits_{n = 1}^{\infty}\dfrac{x^2}{(1 + x^2)^n}\) 对任意的 \(x\) 绝对收敛. 接下来考虑 \(\sum\limits_{k = n + 1}^{n + p}u_k(x)\).

\[ \begin{aligned} \sum\limits_{k = n + 1}^{n + p}u_k(x) & = \sum\limits_{k = n + 1}^{n + p}\dfrac{x^2}{(1 + x^2)^{k}} \\ & = \dfrac{x^2}{(1 + x^2)^{n + 1}}\times\dfrac{1 - \left(\frac{1}{1 + x^2}\right)^p}{1 - \frac{1}{1 + x^2}} \\ & = \dfrac{1 - \left(\frac{1}{1 + x^2}\right)^p}{(1 + x^2)^n} \end{aligned} \]

取 \(n = p = N + 1, x = \sqrt{\sqrt[N + 1]{2} - 1}, \epsilon = \dfrac{1}{4}\), 则

\[ \begin{aligned} \sum\limits_{k = n + 1}^{n + p}u_k(x) & = \dfrac{1 - \left(\frac{1}{1 + x^2}\right)^p}{(1 + x^2)^n} \\ & = \dfrac{1 - \left(\frac{1}{1 + \sqrt[N+1]{2} - 1}\right)^{N+1}}{(1 + \sqrt[N+1]{2} - 1)^{N+1}} \\ & = \dfrac{1 - \frac{1}{2}}{2} \\ & = \dfrac{1}{4} = \epsilon \end{aligned} \]

因此, 级数 \(\sum\limits_{n = 1}^{\infty}\dfrac{x^2}{(1 + x^2)^n}\) 在 \((-\infty, +\infty)\) 上非一致收敛.

第 6 题

证明: 函数 \(f(x) = \sum\limits_{n = 1}^{\infty}ne^{-nx}\) 在 \((0, +\infty)\) 上连续, 进一步证明在 \((0, +\infty)\) 上可微.

证明

首先, \(u_n(x) = ne^{-nx}\) 在 \(x \in (0, +\infty)\) 上连续可导, \(u'_n(x) = -n^2e^{-nx}\). 其次, \(\forall \delta > 0\), \(-\sum\limits_{n = 1}^{\infty}u'_n = \sum\limits_{n = 1}^{\infty}n^2e^{-nx} < \sum\limits_{n = 1}^{\infty}n^2e^{-n\delta}\), 而 \(\sum\limits_{n = 1}^{\infty}n^2e^{-n\delta}\) 收敛. 所以由 Weierstrass 判别法知 \(\sum\limits_{n=1}^{\infty}u'_n(x)\) 在 \((\delta, +\infty)\) 上一致收敛. 由于 \(\delta\) 可以任意小, 故 \(\sum\limits_{n=1}^{\infty}u'_n(x)\) 在 \(x \in (0, +\infty)\) 上一致收敛. 最后, \(\exists x = x_0 = 1\) 时 \(f(1) = \sum\limits_{n = 1}^{\infty}\dfrac{n}{e^n}\) 收敛. 因此, 函数 \(f(x)\) 在 \((0, +\infty)\) 上连续, \(f(x) = \sum\limits_{n = 1}^{\infty}ne^{-nx}\) 在 \((0, +\infty)\) 在 \((0, +\infty)\) 上可微.