\begin{align*}
\lim\limits_{n\to+\infty}\left(\dfrac{1}{n}+\dfrac{1}{n+1}+\cdots+\dfrac{1}{3n+2}\right)=\ln3
\end{align*}
看来多走几步就出来了
\begin{align*}
\int_0^{\tfrac{1}{n}} \dfrac{1}{1+x}{\mathrm{d}}x=\ln\left(\dfrac{n+1}{n}\right)<\dfrac{1}{n}&<\ln\left(\dfrac{n}{n-1}\right)=\int_{-\tfrac{1}{n}}^0 \dfrac{1}{1+x}{\mathrm{d}}x\\
\int_{\tfrac{1}{n}}^{\tfrac{2}{n}} \dfrac{1}{1+x}{\mathrm{d}}x=\ln\left(\dfrac{n+2}{n+1}\right)<\dfrac{1}{n+1}&<\ln\left(\dfrac{n+1}{n}\right)=\int_0^{\tfrac{1}{n}} \dfrac{1}{1+x}{\mathrm{d}}x\\
\vdots\quad\\
\int_2^{\tfrac{2n+1}{n}} \dfrac{1}{1+x}{\mathrm{d}}x=\ln\left(\dfrac{3n+1}{3n}\right)<\dfrac{1}{3n}&<\ln\left(\dfrac{3n}{3n-1}\right)=\int_{\tfrac{2n-1}{n}}^2 \dfrac{1}{1+x}{\mathrm{d}}x\\
\int_{\tfrac{2n+1}{n}}^{\tfrac{2n+2}{n}} \dfrac{1}{1+x}{\mathrm{d}}x=\ln\left(\dfrac{3n+2}{3n+1}\right)<\dfrac{1}{3n+1}&<\ln\left(\dfrac{3n+1}{3n}\right)=\int_{\tfrac{2n}{n}}^{\tfrac{2n+1}{n}} \dfrac{1}{1+x}{\mathrm{d}}x\\
\int_{\tfrac{2n+2}{n}}^{\tfrac{2n+3}{n}} \dfrac{1}{1+x}{\mathrm{d}}x=\ln\left(\dfrac{3n+3}{3n+2}\right)<\dfrac{1}{3n+2}&<\ln\left(\dfrac{3n+2}{3n+1}\right)=\int_{\tfrac{2n+1}{n}}^{\tfrac{2n+2}{n}} \dfrac{1}{1+x}{\mathrm{d}}x\\
\end{align*} |
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发表于 2020-9-9 23:15
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发表于 2020-11-23 16:16
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