极限练习题（一）
作者：互联网
\(\mathbf{Qn1}\quad \displaystyle{\lim_{N\rightarrow\infty} \sum_{n=1}^N \left(n\ln\frac{2n+1}{2n-1}-1\right) }\)
\(\mathbf{Sol}\)
\[\begin{aligned} \lim_{N\rightarrow\infty} \sum_{n=1}^N \left(n\ln\frac{2n+1}{2n-1}-1\right)=&\lim_{N\rightarrow\infty}\ln\left(\text e^{-N}\prod_{n=1}^N \left(\frac{2n+1}{2n-1}\right)^n \right)\\ =&\lim_{N\rightarrow\infty}\ln\left(\text e^{-N}\frac{(2N+1)^N2^NN!}{(2N)!}\right)\\ =&\lim_{N\rightarrow\infty}\ln\left((1+\frac{1}{2N})^N\cdot\frac{2^{2N}N^NN!}{\text e^N(2N)!}\right)\\ =&\lim_{N\rightarrow\infty}\ln\left((1+\frac{1}{2N})^N\cdot\frac{2^{2N}N^N\sqrt{2\pi N}(\frac{N}{\text e})^N}{\text e^N\sqrt{4\pi N}(\frac{2N}{\text e})^{2N}}\right)\\ =&\lim_{N\rightarrow\infty}\ln\left(\sqrt\frac{\text e}{2}\right)\\ =&\frac{1-\ln2}{2} \end{aligned} \]标签：练习题,right,frac,text,极限,2N,rightarrow,left 来源： https://www.cnblogs.com/Arcticus/p/16308437.html