자연계열 (오후1) 2번

(2-1) $\displaystyle a=2n,b=2n+3$로 택하자. 제시문 (가)에 있는 부등식 $\displaystyle \frac1b<\frac{\ln b-\ln a}{b-a}<\frac1a$에 대입하면$\displaystyle \begin{aligned}\frac1{2n+3}<\frac{\ln(2n+3)-\ln(2n)}3<\frac1{2n}\quad&\Rightarrow\quad\frac1{2n+3}<\frac13\ln\left(1+\frac3{2n}\right)<\frac1{2n}\\&\Rightarrow\quad\frac{3n}{2n+3}<\ln\left(1+\frac3{2n}\right)^n<\frac32\end{aligned}$

마지막 부등식에 $\displaystyle \lim_{n\to\infty}$을 취하면 제시문 (나)와 제시문 (다)에 의하여$\displaystyle \lim_{n\to\infty}\ln\left(1+\frac3{2n}\right)^n=\ln\left(\lim_{n\to\infty}\left(1+\frac3{2n}\right)^n\right)=\frac32$이다.따라서 $\displaystyle \lim_{n\to\infty}\left(1+\frac3{2n}\right)^n=e^{\frac32}=e\sqrt e$이다.

(2-2)(a) 위 그림에서 $\displaystyle OA=\frac1{\sqrt3}$, $\displaystyle AB=\sqrt{1+OA^2}=\sqrt{1+\frac13}=\frac2{\sqrt3}=BC=a_2$$\displaystyle 2\theta_2=\theta_1=30^{\circ}$이므로 $\displaystyle \theta_2=15^{\circ}$이다. 따라서$\displaystyle \sin15^{\circ}=\frac{OA}{CA},\ OA^2=\frac13,\ CA^2=OA^2+(a_1+a_2)^2=\frac13+\left(1+\frac2{\sqrt3}\right)^2=\frac83+\frac4{\sqrt3}=\frac{8+4\sqrt3}3$$\displaystyle \sin^215^{\circ}=\frac{OA^2}{CA^2}=\frac{2-\sqrt3}4$

(2-2)(b) 먼저 $\displaystyle \theta_n=\frac{\theta_1}{2^{n-1}}=\frac\pi{3\times2^n}$, $\displaystyle \lim_{n\to\infty}\theta_n=0$, $\displaystyle \lim_{n\to\infty}\frac{\sin\theta_n}{\theta_n}=1$이다.또한 $\displaystyle OA=\frac1{\sqrt3}$, $\displaystyle a_{n+1}=\frac{OA}{\sin\theta_n}=\frac1{\sqrt3\sin\theta_n}$이므로$\displaystyle \begin{aligned}\lim_{n\to\infty}\frac{2^n}{a_{n+1}}&=\lim_{n\to\infty}\frac{2^n}{\frac1{\sqrt3\sin\theta_n}}=\lim_{n\to\infty}\left(2^n\sqrt3\frac{\sin\theta_n}{\theta_n}\theta_n\right)\\&=\lim_{n\to\infty}\left(2^n\sqrt3\frac{\sin\theta_n}{\theta_n}\frac\pi{3\times2^n}\right)=\frac{\sqrt3}3\pi=\frac\pi{\sqrt3}\end{aligned}$