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几个和式极限的对比

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青青子衿 posted 2019-6-13 23:15 |Read mode
\begin{align*}
\lim\limits_{n\to+\infty}\sum\limits_{k=1}^{n}\dfrac{n}{n^2+k^2}&=\lim\limits_{n\to+\infty}\dfrac{1}{n}\sum\limits_{k=1}^{n}\dfrac{1}{1+\left(\tfrac{k}{n}\right)^2}=\int^{1}_0\dfrac{1}{1+x^2}\mathrm{d}x=\dfrac{\pi}{4}\\
\lim\limits_{n\to+\infty}\sum\limits_{k=1}^{mn}\dfrac{n}{n^2+k^2}&=\lim\limits_{n\to+\infty}\dfrac{1}{n}\sum\limits_{k=1}^{mn}\dfrac{1}{1+\left(\tfrac{k}{n}\right)^2}=\int^{m}_0\dfrac{1}{1+x^2}\mathrm{d}x=\arctan\left(m\right)\\
\lim\limits_{n\to+\infty}\sum\limits_{k=1}^{n^2}\dfrac{n}{n^2+k^2}&=\lim\limits_{n\to+\infty}\dfrac{1}{n}\sum\limits_{k=1}^{n^2}\dfrac{1}{1+\left(\tfrac{k}{n}\right)^2}=\int^{+\infty}_0\dfrac{1}{1+x^2}\mathrm{d}x=\dfrac{\pi}{2}\\
\lim\limits_{n\to+\infty}\sum\limits_{k=1}^{n^3}\dfrac{n}{n^2+k^2}&=\lim\limits_{n\to+\infty}\dfrac{1}{n}\sum\limits_{k=1}^{n^3}\dfrac{1}{1+\left(\tfrac{k}{n}\right)^2}=\int^{+\infty}_0\dfrac{1}{1+x^2}\mathrm{d}x=\dfrac{\pi}{2}\\
\lim\limits_{n\to+\infty}\sum\limits_{k=1}^{+\infty}\dfrac{n}{n^2+k^2}&=\lim\limits_{n\to+\infty}\dfrac{1}{n}\sum\limits_{k=1}^{+\infty}\dfrac{1}{1+\left(\tfrac{k}{n}\right)^2}=\lim\limits_{n\to+\infty}\left(\frac{\pi}{2}\coth\left(n\pi\right)-\frac{1}{2n}\right)=\dfrac{\pi}{2}\\
\end{align*}

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