자연계열 3번

3-1. [풀이]$\displaystyle \frac{s_1(m)}{r_1(m)}=\frac{{}_{3m}\mathrm{C}_m}{{}_{2m}\mathrm{C}_m}=\frac{\frac{(3m)!}{m!(2m)!}}{\frac{(2m)!}{m!m!}}=\frac{(2m+1)(2m+2)\cdots(2m+m)}{(m+1)(m+2)\cdots(m+m)}$$\displaystyle \begin{aligned}\ln\frac{s_1(m)}{r_1(m)}&=\ln\frac{{}_{3m}\mathrm{C}_m}{{}_{2m}\mathrm{C}_m}=\sum_{k=1}^m\ln(2m+k)-\sum_{k=1}^m\ln(m+k)\\&=\sum_{k=1}^m\ln m\left(2+\frac km\right)-\sum_{k=1}^m\ln m\left(1+\frac km\right)\\&=\sum_{k=1}^m\ln\left(2+\frac km\right)-\sum_{k=1}^m\ln\left(1+\frac km\right)\end{aligned}$$\displaystyle \begin{aligned}\lim_{m\to\infty}\frac{(2m)^{m-1}}{h_1(m)}\ln\frac{s_1(m)}{r_1(m)}&=\lim_{m\to\infty}\frac1{2m}\left(\sum_{k=1}^m\ln\left(2+\frac km\right)-\sum_{k=1}^m\ln\left(1+\frac km\right)\right)=\frac12\left[\int_2^3\ln x\,dx-\int_1^2\ln x\,dx\right]\\&=\frac12[3\ln3-4\ln2]\end{aligned}$

3-2. [풀이]$\displaystyle \frac{s_2(m)}{r_2(m)}=\frac{{}_{4m}\mathrm{C}_{2m}}{{}_{3m}\mathrm{C}_{2m}}=\frac{\frac{(4m)!}{2m!(2m)!}}{\frac{(3m)!}{(2m)!m!}}=\frac{(2m+1)\cdots(2m+2m)}{(m+1)\cdots(m+2m)}$$\displaystyle \begin{aligned}\ln\frac{s_2(m)}{r_2(m)}&=\ln\frac{{}_{4m}\mathrm{C}_{2m}}{{}_{3m}\mathrm{C}_{2m}}=\sum_{k=1}^{2m}\ln(2m+k)-\sum_{k=1}^{2m}\ln(m+k)\\&=\sum_{k=1}^{2m}\ln\left(2+\frac km\right)-\sum_{k=1}^{2m}\ln\left(1+\frac km\right)\end{aligned}$$\displaystyle \frac{s_3(m)}{r_3(m)}=\frac{{}_{5m}\mathrm{C}_{3m}}{{}_{4m}\mathrm{C}_{3m}}=\frac{\frac{(5m)!}{3m!(2m)!}}{\frac{(4m)!}{(3m)!m!}}=\frac{(2m+1)\cdots(2m+3m)}{(m+1)\cdots(m+3m)}$$\displaystyle \ln\frac{s_3(m)}{r_3(m)}=\ln\frac{{}_{5m}\mathrm{C}_{3m}}{{}_{4m}\mathrm{C}_{3m}}=\sum_{k=1}^{3m}\ln\left(2+\frac km\right)-\sum_{k=1}^{3m}\ln\left(1+\frac km\right)$$\displaystyle \vdots$$\displaystyle \frac{s_{10}(m)}{r_{10}(m)}=\frac{{}_{12m}\mathrm{C}_{10m}}{{}_{11m}\mathrm{C}_{10m}}=\frac{\frac{(12m)!}{10m!(2m)!}}{\frac{(11m)!}{(10m)!m!}}=\frac{(2m+1)\cdots(2m+10m)}{(m+1)\cdots(m+10m)}$$\displaystyle \ln\frac{s_{10}(m)}{r_{10}(m)}=\ln\frac{{}_{12m}\mathrm{C}_{10m}}{{}_{11m}\mathrm{C}_{10m}}=\sum_{k=1}^{10m}\ln\left(2+\frac km\right)-\sum_{k=1}^{10m}\ln\left(1+\frac km\right)$$\displaystyle \begin{aligned}&\lim_{m\to\infty}\frac{(2m)^{m-1}}{h_1(m)}\ln\frac{s_1(m)\cdots s_{10}(m)}{r_1(m)\cdots r_{10}(m)}\\&=\frac12\left[\int_2^3\ln x\,dx-\int_1^2\ln x\,dx+\int_2^4\ln x\,dx-\int_1^3\ln x\,dx+\int_2^5\ln x\,dx-\int_1^4\ln x\,dx+\cdots+\int_2^{11}\ln x\,dx-\int_1^{10}\ln x\,dx+\int_2^{12}\ln x\,dx-\int_1^{11}\ln x\,dx\right]\\&=\frac12\left[\int_2^3\ln x\,dx-\int_1^2\ln x\,dx+\int_3^4\ln x\,dx-\int_1^2\ln x\,dx+\cdots+\int_{11}^{12}\ln x\,dx-\int_1^2\ln x\,dx\right]\\&=\frac12\left[\int_2^{12}\ln x\,dx-10\int_1^2\ln x\,dx\right]\\&=6\ln3+\ln2\end{aligned}$