자연계열 (오후2) 2번

1. 함수 $\displaystyle f(x)=\sin x$는 미분 가능하므로,평균값 정리에 의해 $\displaystyle a<x<b$인 실수 $\displaystyle x$에 대하여 $\displaystyle \sin b-\sin x=(\cos\alpha)(b-x)$인 $\displaystyle \alpha\in(x,b)$가 존재한다.또한, $\displaystyle f^{\prime}(x)=\cos x$는 구간 $\displaystyle [0,\pi]$에서 감소하므로, $\displaystyle \cos a\ge\cos\alpha\ge\cos b$가 성립한다.그런데, $\displaystyle b-x>0$이므로, $\displaystyle (b-x)\cos a\ge(b-x)\cos\alpha=\sin b-\sin x\ge(b-x)\cos b$이다.

각 변을 $\displaystyle a$에서 $\displaystyle b$까지 적분하면, $\displaystyle \int_a^b(b-x)\cos b\,dx\le\int_a^b(\sin b-\sin x)\,dx\le\int_a^b(b-x)\cos a\,dx$이다.그러므로 $\displaystyle \frac12(b-a)^2\cos b\le\int_a^b(\sin b-\sin x)\,dx\le\frac12(b-a)^2\cos a$이다.

2. $\displaystyle \int e^{-x}\cos x\,dx=-e^{-x}\cos x-\int e^{-x}\sin x\,dx=-e^{-x}\cos x-\left[-e^{-x}\sin x+\int e^{-x}\cos x\,dx\right]$$\displaystyle \therefore\int e^{-x}\cos x\,dx=\frac12e^{-x}(\sin x-\cos x).$① $\displaystyle \int_{\left(k-\frac12\right)\pi}^{\left(k+\frac12\right)\pi}e^{-x}\cos x\,dx=\Biggl[\frac12e^{-x}(\sin x-\cos x)\Biggr]_{\left(k-\frac12\right)\pi}^{\left(k+\frac12\right)\pi}=(-1)^k\frac12\left(e^{-\left(k+\frac12\right)\pi}+e^{-\left(k-\frac12\right)\pi}\right)$

② $\displaystyle \begin{aligned}a_n&=\int_{\left(k-\frac12\right)\pi}^{\left(k+\frac12\right)\pi}|e^{-x}\cos x|\,dx\\&=\begin{cases}\int_{\left(k-\frac12\right)\pi}^{\left(k+\frac12\right)\pi}e^{-x}\cos x\,dx,&k=\text{짝수}\\-\int_{\left(k-\frac12\right)\pi}^{\left(k+\frac12\right)\pi}e^{-x}\cos x\,dx,&k=\text{홀수}\end{cases}\\&=\frac12\left(e^{-\left(k+\frac12\right)\pi}+e^{-\left(k-\frac12\right)\pi}\right)=\frac12\left(e^{-\left(k-\frac12\right)\pi}+e^{-\left(k+\frac12\right)\pi}\right)\end{aligned}$따라서 급수의 합은 다음과 같다.$\displaystyle \begin{aligned}\sum_{n=1}^\infty a_n&=\lim_{n\to\infty}\sum_{k=1}^n\frac12\left(e^{-\left(k-\frac12\right)\pi}+e^{-\left(k+\frac12\right)\pi}\right)\\&=\frac12\lim_{n\to\infty}\left[e^{-\frac12\pi}+e^{-\frac32\pi}+e^{-\frac32\pi}+e^{-\frac52\pi}+\cdots+e^{-\left(n-\frac12\right)\pi}+e^{-\left(n+\frac12\right)\pi}\right]\\&=\frac12\lim_{n\to\infty}\left[e^{-\frac12\pi}+2\frac{e^{-\frac32\pi}(1-e^{-(n-1)\pi})}{1-e^{-\pi}}+e^{-\left(n+\frac12\right)\pi}\right]\\&=\frac12e^{-\frac12\pi}+\frac{e^{-\frac32\pi}}{1-e^{-\pi}}=\frac12\frac{e^{-\frac12\pi}+e^{-\frac32\pi}}{1-e^{-\pi}}=\frac{e^{-\frac12\pi}}2\frac{1+e^{-\pi}}{1-e^{-\pi}}\end{aligned}$

3. $\displaystyle b_1=\frac32,\quad b_k-b_{k-1}=\frac1{k+1}$이다.$\displaystyle \begin{aligned}\sum_{k=1}^n\frac{b_k}{k(k+1)}&=\sum_{k=1}^n\left(\frac{b_k}k-\frac{b_k}{k+1}\right)=\left(b_1-\frac{b_1}2\right)+\left(\frac{b_2}2-\frac{b_2}3\right)+\cdots+\left(\frac{b_n}n-\frac{b_n}{n+1}\right)\\&=b_1+\frac12(b_2-b_1)+\frac13(b_3-b_2)+\cdots+\frac1n(b_n-b_{n-1})-\frac{b_n}{n+1}\\&=\frac32+\left(\frac12-\frac13\right)+\left(\frac13-\frac14\right)+\cdots+\left(\frac1n-\frac1{n+1}\right)-\frac{b_n}{n+1}\\&=2-\frac1{n+1}-\frac{b_n}{n+1}.\end{aligned}$

부등식 $\displaystyle \ln(n+1)=\int_1^{n+1}\frac1x\,dx\le b_n=\sum_{k=1}^{n+1}\frac1k\le1+\int_1^{n+1}\frac1x\,dx=1+\ln(n+1)$이 성립하므로,각 변을 $\displaystyle n+1$로 나누고 극한 $\displaystyle \lim_{n\to\infty}$을 취하면, $\displaystyle 0=\lim_{n\to\infty}\frac{\ln(n+1)}{n+1}\le\lim_{n\to\infty}\frac{b_n}{n+1}\le\lim_{n\to\infty}\left(\frac1{n+1}+\frac{\ln(n+1)}{n+1}\right)=0$이다.

따라서 $\displaystyle \lim_{n\to\infty}\frac1{n+1}=\lim_{n\to\infty}\frac{b_n}{n+1}=0$이므로, $\displaystyle \lim_{n\to\infty}\sum_{k=1}^n\frac{b_k}{k(k+1)}=2$이다.