一个极限的小问题

青青子衿

\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*}

青青子衿

\begin{align*}
\lim\limits_{n\to+\infty}\left(\dfrac{1}{n}+\dfrac{1}{n+1}+\cdots+\dfrac{1}{3n+2}\right)&=\lim\limits_{n\to+\infty}\dfrac{1}{n}\sum_{k=0}^{2n+2}\dfrac{1}{1+\frac{k}{n}}\\
&=\lim\limits_{n\to+\infty}\dfrac{1}{n}\sum_{k=1}^{2n}\dfrac{1}{1+\frac{k}{n}}\\
&=2\lim\limits_{n\to+\infty}\dfrac{1}{2n}\sum_{k=1}^{2n}\dfrac{1}{1+2\left(\frac{k}{2n}\right)}\\
&=2\lim\limits_{n\to+\infty}\dfrac{1}{m}\sum_{k=1}^{m}\dfrac{1}{1+2\left(\frac{k}{m}\right)}\\
&=2\int_0^1\dfrac{1}{1+2x}\mathrm{d}x=\int_0^1\dfrac{1}{1+2x}\mathrm{d}(2x)\\
&=\int_0^2\dfrac{1}{1+u}\mathrm{d}u=\ln3
\end{align*}

$ \displaystyle\frac{1}{n}\sum_{k=1}^{3n}\arctan\left(1+\frac{2k}{n}\right)=\int_0^3\arctan\left(1+2t\right)\mathrm{d}t $

zhcosin

借助欧拉常数$c$，有
\[ \sum_{k=1}^n \frac{1}{k} = \ln{n} +c + o_n(1) \]
及
\[ \sum_{k=1}^{3n+2} \frac{1}{k} = \ln{(3n+2)} +c + o_{3n+2}(1) \]
所以
\[ \sum_{k=n}^{3n+2} \frac{1}{k} = \ln{(3n+2)} -  \ln{n} + o(1) = \ln{\left( 3+\frac{2}{n} \right)} +o(1) \]
即得结论.