回復 34# shmilypon 的帖子
偵錯題第二題
問題：求 \(\mathop {\lim }\limits_{n \to \infty } \sum\limits_{k = 1}^n {\frac{1}{n}} \times \frac{1}{{{k^2}}}\)=?
錯解：作區間\(\left[ {0,1} \right]\)的\(n\)等分分割，則\(\sum\limits_{k = 1}^n {\frac{1}{n}} \times \frac{1}{{{k^2}}} = \int_0^1 {\frac{1}{{{x^2}}}dx = \left( {\frac{{ - 1}}{x}} \right)} \left| {_0^1} \right.\) 故極限不存在
解答
\(\mathop {\lim }\limits_{n \to \infty } \sum\limits_{k = 1}^n {\frac{1}{n}} \times \frac{1}{{{k^2}}} = \mathop {\lim }\limits_{n \to \infty } \frac{1}{n}\sum\limits_{k = 1}^n {\frac{1}{{{k^2}}}} = \mathop {\lim }\limits_{n \to \infty } \left\{ {\frac{1}{n} \times \left( {\frac{1}{{{1^2}}} + \frac{1}{{{2^2}}} + \frac{1}{{{3^2}}} + \cdots + \frac{1}{{{n^2}}}} \right)} \right\}\)
(1) 證明 \(\mathop {\lim }\limits_{n \to \infty } \left( {\frac{1}{{{1^2}}} + \frac{1}{{{2^2}}} + \frac{1}{{{3^2}}} + \cdots + \frac{1}{{{n^2}}}} \right) = \frac{{{\pi ^2}}}{6}\)
令 \(f\left( x \right)=\sin x\)求 \(\sin x\) 在 \(x=0\) 的泰勒展開式
\(f\left( x \right)=f\left( 0 \right)+\frac{{f}'\left( 0 \right)}{1!}x+\frac{{f}''\left( 0 \right)}{2!}{{x}^{2}}+\frac{{f}'''\left( 0 \right)}{3!}{{x}^{3}}+\cdots +\frac{{{f}^{(n)}}\left( \delta \right)}{n!}{{x}^{n}}\ \ \ \delta \in \left( -\varepsilon ,\varepsilon \right)\)
\(\begin{align}
& f\left( 0 \right)=\sin 0=0\ ,\ {f}'\left( x \right)\ =\cos x\ ,\ {f}'\left( 0 \right)=\cos 0=1\ ,\ {f}''\left( x \right)=\sin x \\
& {f}''\left( 0 \right)=0\ ,\ {f}'''\left( x \right)=-\cos x\ ,\ {f}'''\left( 0 \right)=-1\ ,\ {{f}^{(4)}}\left( x \right)=\sin x\ ,\ {{f}^{(4)}}\left( 0 \right)=0 \\
& {{f}^{(5)}}\left( x \right)=\cos x\ ,\ {{f}^{(5)}}\left( 0 \right)=1\ ,\ \cdots \\
\end{align}\)
(2)
\(\begin{align}
& \sin x=\left\{ \frac{x}{1!}-\frac{{{x}^{3}}}{3!}+\frac{{{x}^{5}}}{5!}-\frac{{{x}^{7}}}{7!}+\frac{{{x}^{9}}}{9!}-\frac{{{x}^{11}}}{11!}+\cdots \right\} \\
& \frac{\sin x}{x}=1-\frac{{{x}^{2}}}{3!}+\frac{{{x}^{4}}}{5!}-\frac{{{x}^{6}}}{7!}+\frac{{{x}^{8}}}{9!}-\frac{{{x}^{10}}}{11!}+\cdots \ \ \ \ \ \ \ \ x\ne 0 \\
\end{align}\)
(3) 當 \(\frac{\sin x}{x}=0\) 方程式的根，顯然是\( \pm \pi , \pm 2\pi , \pm 3\pi , \pm 4\pi , \cdots \;\;\;\;\;\left( {x \ne 0} \right)\)
\(\begin{array}{l}
\Rightarrow \;1 - \frac{{{x^2}}}{{3!}} + \frac{{{x^4}}}{{5!}} - \frac{{{x^6}}}{{7!}} + \frac{{{x^8}}}{{9!}} - \frac{{{x^{10}}}}{{11!}} + \cdots \\
\;\; = \left( {1 - \frac{x}{\pi }} \right) \times \left( {1 + \frac{x}{\pi }} \right) \times \left( {1 - \frac{x}{{2\pi }}} \right) \times \left( {1 + \frac{x}{{2\pi }}} \right) \times \left( {1 - \frac{x}{{3\pi }}} \right) \times \left( {1 + \frac{x}{{3\pi }}} \right) \times \cdots \\
\;\; = \left( {1 - \frac{{{x^2}}}{{{\pi ^2}}}} \right) \times \left( {1 - \frac{{{x^2}}}{{{2^2}{\pi ^2}}}} \right) \times \left( {1 - \frac{{{x^2}}}{{{3^2}{\pi ^2}}}} \right) \times \cdots
\end{array}\)
考慮\({{x}^{2}}\) 項的係數
\(\begin{align}
& \frac{-1}{{{\pi }^{2}}}-\frac{1}{{{2}^{2}}{{\pi }^{2}}}-\frac{1}{{{3}^{2}}{{\pi }^{2}}}-\cdots \cdots =\frac{-1}{3!} \\
& \Rightarrow \frac{1}{{{1}^{2}}}+\frac{1}{{{2}^{2}}}+\frac{1}{{{3}^{2}}}+\cdots \cdots =\frac{{{\pi }^{2}}}{6} \\
\end{align}\)
得證
(4)
\(\underset{n\to \infty }{\mathop{\lim }}\,\sum\limits_{k=1}^{n}{\frac{1}{n}}\times \frac{1}{{{k}^{2}}}=\underset{n\to \infty }{\mathop{\lim }}\,\left\{ \frac{1}{n}\times \left( \frac{1}{{{1}^{2}}}+\frac{1}{{{2}^{2}}}+\frac{1}{{{3}^{2}}}+\cdots +\frac{1}{{{n}^{2}}} \right) \right\}\)
且 \(\underset{n\to \infty }{\mathop{\lim }}\,\frac{1}{n}\) 與\(\underset{n\to \infty }{\mathop{\lim }}\,\left( \frac{1}{{{1}^{2}}}+\frac{1}{{{2}^{2}}}+\cdots +\frac{1}{{{n}^{2}}} \right)\) 極限值皆存在
\(\begin{align}
& \underset{n\to \infty }{\mathop{\lim }}\,\sum\limits_{k=1}^{n}{\frac{1}{n}}\times \frac{1}{{{k}^{2}}}=\underset{n\to \infty }{\mathop{\lim }}\,\left\{ \frac{1}{n}\times \left( \frac{1}{{{1}^{2}}}+\frac{1}{{{2}^{2}}}+\frac{1}{{{3}^{2}}}+\cdots +\frac{1}{{{n}^{2}}} \right) \right\} \\
& \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\underset{n\to \infty }{\mathop{\lim }}\,\frac{1}{n}\times \underset{n\to \infty }{\mathop{\lim }}\,\left( \frac{1}{{{1}^{2}}}+\frac{1}{{{2}^{2}}}+\frac{1}{{{3}^{2}}}+\cdots +\frac{1}{{{n}^{2}}} \right)=0\times \frac{{{\pi }^{2}}}{6} \\
\end{align}\) 收斂到 \(0\)
學生錯誤的原因，\(\frac{1}{n}\) 很容易聯想到黎曼和 \(\left[ 0,1 \right]\) 區間做分割。
\(\Delta x=\frac{1-0}{n}\) ，分割點 \(\left\{ 0,\frac{1}{n},\frac{2}{n},\frac{3}{n},\frac{4}{n},\cdots ,\frac{n}{n} \right\}\)
黎曼和的高是 \(f\left( x \right)=\frac{1}{{{x}^{2}}}\ ,\ f\left( \frac{1}{n} \right)={{n}^{2}}\ ,\ f\left( \frac{2}{n} \right)=\frac{{{n}^{2}}}{4}\ ,\cdots \)
黎曼和寫出來的式子是 \(\underset{n\to \infty }{\mathop{\lim }}\,\sum\limits_{i=1}^{n}{\frac{1}{n}}f\left( \frac{i}{n} \right)\) 。這個題目是 \(\sum\limits_{k=1}^{n}{\frac{1}{{{k}^{2}}}}\) P級數求和
學生把這個題目當成黎曼和分割處理。底的部分 1/n, 是可以的。。
錯誤地方在黎曼和的高是用分割點代入。所以黎曼和寫出來的式子。不是這個題目要求的答案。
[ 本帖最後由 shingjay176 於 2014-5-10 10:00 AM 編輯 ]
