Review Note

平方数列的前n项和\(\begin{aligned}\displaystyle\sum_{k=1}^{n} k^{2}=\end{aligned}\)________.

\(\begin{aligned}\displaystyle\sum_{k=1}^{n} k^{2}=1^{2}+2^{2}+3^{2}+\cdots+n^{2}=\frac{n(n+1)(2 n+1)}{6}\end{aligned}\)

同济教材推导
利用恒等式 \((n+1)^{3}=n^{3}+3 n^{2}+3 n+1\), 得
\(\left\{\begin{array}{l} (n+1)^{3}-n^{3}=3 n^{2}+3 n+1, \\ n^{3}-(n-1)^{3}=3(n-1)^{2}+3(n-1)+1, \\ \cdots \cdots \cdots \cdots \\ 3^{3}-2^{3}=3 \cdot 2^{2}+3 \cdot 2+1, \\ 2^{3}-1^{3}=3 \cdot 1^{2}+3 \cdot 1+1 . \end{array}\right.\)
把这 \(n\) 个等式两端分别相加, 得
\((n+1)^{3}-1=3\left(1^{2}+2^{2}+\cdots+n^{2}\right)+3(1+2+\cdots+n)+n .\)
由于 \(1+2+\cdots+n=\frac{1}{2} n(n+1)\), 代入上式, 得
\(n^{3}+3 n^{2}+3 n=3\left(1^{2}+2^{2}+\cdots+n^{2}\right)+\frac{3}{2} n(n+1)+n .\)
整理后, 得\(1^{2}+2^{2}+\cdots+n^{2}=\frac{1}{6} n(n+1)(2 n+1) .\)"

\(\begin{align}
\lim _{n \rightarrow \infty}\left(1+\frac{1}{n^2}\right)\left(1+\frac{2}{n^2}\right) \cdots\left(1+\frac{n}{n^2}\right)
\end{align}\)

\(\begin{aligned} & \ln \left[\left(1+\frac{1}{n^2}\right)\left(1+\frac{2}{n^2}\right) \cdots\left(1+\frac{n}{n^2}\right)\right] \\ & =\ln \left(1+\frac{1}{n^2}\right)+\ln \left(1+\frac{2}{n^2}\right)+\cdots+\ln \left(1+\frac{n}{n^2}\right) \\ & =\sum_{k=1}^n \ln \left(1+\frac{k}{n^2}\right) \\ & \frac{1}{n^6} \leq\left(\frac{k}{n^2}\right)^3 \leq \frac{n^3}{n^6}=\frac{1}{n^3} \Rightarrow\left(\frac{k}{n^2}\right)^3=o\left(\frac{1}{n^2}\right) \\ & \ln \left(1+\frac{k}{n^2}\right)=\frac{k}{n^2}-\frac{1}{2}\left(\frac{k}{n^2}\right)^2+o\left(\frac{1}{n^2}\right) \\ & \lim _{n \rightarrow \infty} \sum_{k=1}^n \ln \left(1+\frac{k}{n^2}\right)=\lim _{n \rightarrow \infty} \sum_{k=1}^n\left[\frac{k}{n^2}-\frac{1}{2}\left(\frac{k}{n^2}\right)^2+o\left(\frac{1}{n^2}\right)\right] \\ & =\lim _{n \rightarrow \infty} \sum_{k=1}^n \frac{k}{n^2}-\frac{1}{2} \lim _{n \rightarrow \infty} \sum_{k=1}^n \frac{k^2}{n^4}+\lim _{n \rightarrow \infty} n * o\left(\frac{1}{n^2}\right) \\ & =\lim _{n \rightarrow \infty} \frac{1}{n^2} * \frac{n(n+1)}{2}-\frac{1}{2} \lim _{n \rightarrow \infty} \frac{1}{n^4} * \frac{n(n+1)(2 n+1)}{6}+0 \\ & =\frac{1}{2} \quad \Rightarrow I=e^{\frac{1}{2}}=\sqrt{e}\end{aligned}\)