竞赛好题暑假练习2
练习 竞赛 暑假 好题
2023-06-13 09:13:16 时间
利用积分和级数来求一道极限问题
求极限
\begin{align*}
\lim\limits_{
\begin{subarray}{l}
m \to \infty\
n \to \infty\
\end{subarray}
}\sum\limits_{i=1}^{m} \sum\limits_{j=1}^{n} \dfrac{(-1)^{i+j}}{i+j}
\end{align*}解：由于
\displaystyle\int_{-1}^{0}x^{i+j-1}dx=\dfrac{1}{i+j}x^{i+j}|_{-1}^{0}=-\dfrac{(-1)^{i+j}}{i+j}，所以部分和
\displaystyle \begin{align*}S_{m,n} &=\sum_{i=1}^{m}\sum_{j=1}^{n}\dfrac{(-1)^{i+j}}{i+j}=-\sum_{i=1}^{m}\sum_{j=1}^{n}\int_{-1}^{0}x^{i+j-1}dx \\&=-\sum_{i=1}^{m}(\int_{-1}^{0}x^{i+1-1}dx+\int_{-1}^{0}x^{i+2-1}dx+dotsb+\int_{-1}^{0}x^{i+n-1}dx)\\&=-\sum_{i=1}^{m}\int_{-1}^{0}(x^{i}+x^{i+1}+x^{i+2}+\dotsb+x^{i+n-1}dx)\\&=-\sum_{i=1}^{m}\int_{-1}^{0}\dfrac{x^{i}(1-x^{n})}{1-x}dx=-\sum_{i=1}^{m}\int_{-1}^{0}\dfrac{x^{i}-x^{i+n}}{1-x}dx\\ &=-\int_{-1}^{0}\dfrac{(x-x^{1+n}+(x^2-x^{2+n})+\dotsb+(x^m-x^{m+n}))}{1-x}dx\\ &=-\int_{-1}^{0}\dfrac{\dfrac{x(1-x^{n})}{1-x}}{1-x}dx+\int_{-1}^{0}\dfrac{\dfrac{x^{n+1}(1-x^m)}{1-x}}{1-x}dx\\&=\int_{-1}^{0}\dfrac{x-x^{m+1}}{(1-x)^2}dx+\int_{-1}^{0}\dfrac{x^{n+m+1}}{(1-x)^2}\end{align*}
而当
x\in[-1,0]时，有
1\leq(1-x)^2\geq 4，所以当
l\rightarrow +\infty时，有
\displaystyle \dfrac{1}{4}\int_{-1}^{0}x^{l}dx \leq \int_{-1}^{0}\dfrac{x^l}{(1-x)^2}dx \leq \int_{-1}^{0}x^{l}dx，而
\displaystyle\int_{-1}^{0}x^ldx=\dfrac{1}{l+1}x^{l+1}|_{-1}^{0}=\dfrac{(-1)^{l+2}}{^{l+1}}\quad l\rightarrow 0所以原式
\displaystyle \begin{align*}&=\lim\limits_{m \rightarrow +\infty \\ n\rightarrow +\infty}S_{m,n}=-\int_{-1}^{0}\dfrac{x}{(1-x)^2}\\&=\int_{-1}^{0}\dfrac{1}{(1-x)^2}dx-\int_{-1}^{0}\dfrac{1}{1-x}dx=\ln 2-\dfrac{1}{2}\end{align*}
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