자연계열 II 3번

【3-1】함수 $\displaystyle h(x)$의 부정적분을 $\displaystyle F(x)$라 하자. 각각의 자연수 $\displaystyle i$에 대하여, 평균값 정리에 의해$\displaystyle h(c_i)=\dfrac{F(i+1)-F(i)}{(i+1)-i}=F(i+1)-F(i)$를 만족하는 $\displaystyle c_i$가 열린구간 $\displaystyle (i,i+1)$에 존재한다.함수 $\displaystyle h(x)$는 구간 $\displaystyle [m,\infty)$에서 감소하므로 각각의 자연수 $\displaystyle i$ ($\displaystyle i\ge m$)에 대하여$\displaystyle h(i+1)<h(c_i)<h(i)$가 성립한다. 따라서$\displaystyle \sum_{i=m+1}^{n+1}h(i)<F(n+1)-F(m)<\sum_{i=m}^nh(i)$이다. 한편, 치환적분법에 의해

$\displaystyle F(n+1)-F(m)=\int_m^{n+1}f(g(x))g^{\prime}(x)\,dx=\int_{g(m)}^{g(n+1)}f(u)\,du$이다. 따라서,$\displaystyle \sum_{i=m+1}^{n+1}h(i)<\int_{g(m)}^{g(n+1)}f(x)\,dx<\sum_{i=m}^nh(i)$이다.

【3-2】$\displaystyle f(x)=2^{\frac1x}$, $\displaystyle g(x)=\dfrac1{x+3}$이라 두자.그러면 $\displaystyle h(x)=f(g(x))g^{\prime}(x)=\dfrac{-2^{x+3}}{(x+3)^2}$이고 $\displaystyle \sum_{i=1}^nh(i)=\sum_{i=1}^n\dfrac{-2^{i+3}}{(i+3)^2}$이다.한편, $\displaystyle \sum_{i=1}^1h(i)=-\dfrac{2^4}{4^2}$, $\displaystyle \sum_{i=1}^3h(i)=-\left(\dfrac{2^4}{4^2}+\dfrac{2^5}{5^2}+\dfrac{2^6}{6^2}\right)$, $\displaystyle \sum_{i=1}^4h(i)=-\left(\dfrac{2^4}{4^2}+\dfrac{2^5}{5^2}+\dfrac{2^6}{6^2}+\dfrac{2^7}{7^2}\right)$이다.구간 $\displaystyle (0,\infty)$에서 함수 $\displaystyle f(x)$는 연속이며 함수 $\displaystyle g(x)$는 미분가능이며 $\displaystyle g^{\prime}(x)$는 연속이다.또한, $\displaystyle x>1$일 때 $\displaystyle h^{\prime}(x)=-\dfrac{2^{x+3}\{(x+3)\ln2-2\}}{(x+3)^3}<0$이므로 함수 $\displaystyle h(x)$는 구간 $\displaystyle [1,\infty)$에서 감소한다.

따라서, 【3-1】에 의해$\displaystyle -\left(\dfrac{2^5}{5^2}+\dfrac{2^6}{6^2}+\dfrac{2^7}{7^2}\right)=\sum_{i=2}^4h(i)<\int_{\frac14}^{\frac17}f(x)\,dx<\sum_{i=1}^3h(i)=-\left(\dfrac{2^4}{4^2}+\dfrac{2^5}{5^2}+\dfrac{2^6}{6^2}\right)$이므로 $\displaystyle 2.03\approx\dfrac{2^3}{4^2}+\dfrac{2^4}{5^2}+\dfrac{2^5}{6^2}<\dfrac12\int_{\frac17}^{\frac14}2^{\frac1x}\,dx<\dfrac{2^4}{5^2}+\dfrac{2^5}{6^2}+\dfrac{2^6}{7^2}\approx2.84$이다.따라서, 정답은 $\displaystyle 2$이다.

【3-3】$\displaystyle f(x)=-\ln x$, $\displaystyle g(x)=\dfrac{x^2+1}{x+1}$라 두자.그러면 $\displaystyle h(x)=f(g(x))g^{\prime}(x)=\left\{-\ln\left(\dfrac{x^2+1}{x+1}\right)\right\}\left\{\dfrac{x^2+2x-1}{(x+1)^2}\right\}$이고$\displaystyle \sum_{i=1}^nh(i)=-\sum_{i=1}^n\left[\left\{\ln\left(\dfrac{i^2+1}{i+1}\right)\right\}\left\{\dfrac{i^2+2i-1}{(i+1)^2}\right\}\right]$이다.구간 $\displaystyle (0,\infty)$에서 함수 $\displaystyle f(x)$는 연속이며 함수 $\displaystyle g(x)$는 미분가능이며 $\displaystyle g^{\prime}(x)$는 연속이다.$\displaystyle x>1$일 때 $\displaystyle h^{\prime}(x)=-\left[\left\{\dfrac{x+1}{x^2+1}\right\}\left\{\dfrac{x^2+2x-1}{(x+1)^2}\right\}^2+\left\{\ln\dfrac{x^2+1}{x+1}\right\}\left\{\dfrac4{(x+1)^3}\right\}\right]<0$이므로 함수 $\displaystyle h(x)$는 구간 $\displaystyle [1,\infty)$에서 감소한다.$\displaystyle h(1)=0$이며 문제 【3-1】에 의해 $\displaystyle n\ge1$일 때 아래의 부등식을 얻을 수 있다.$\displaystyle -\sum_{i=1}^{n+1}\left[\left\{\ln\left(\dfrac{i^2+1}{i+1}\right)\right\}\left\{\dfrac{i^2+2i-1}{(i+1)^2}\right\}\right]<\int_1^{\frac{n^2+2n+2}{n+2}}(-\ln x)\,dx<-\sum_{i=1}^n\left[\left\{\ln\left(\dfrac{i^2+1}{i+1}\right)\right\}\left\{\dfrac{i^2+2i-1}{(i+1)^2}\right\}\right]$

따라서, $\displaystyle n\ge1$일 때$\displaystyle a_n=-\dfrac4{(n+1)\ln(n+1)}\sum_{k=1}^{n+1}\left[\left\{\ln\left(\dfrac{k^2+1}{k+1}\right)\right\}\left\{\dfrac{k^2+2k-1}{(k+1)^2}\right\}\right]<\dfrac4{(n+1)\ln(n+1)}\int_1^{\frac{n^2+2n+2}{n+2}}(-\ln x)\,dx$이고 $\displaystyle n\ge2$일 때$\displaystyle \dfrac4{n\ln n}\int_1^{\frac{n^2+2n+2}{n+2}}(-\ln x)\,dx<-\dfrac4{n\ln n}\sum_{k=1}^n\left[\left\{\ln\left(\dfrac{k^2+1}{k+1}\right)\right\}\left\{\dfrac{k^2+2k-1}{(k+1)^2}\right\}\right]=a_{n-1}$이다. 부분적분법에 의해 $\displaystyle \int(-\ln x)\,dx=-x\ln x+\int1\,dx=-x(\ln x-1)+C$이다. ($\displaystyle C$는 적분상수)따라서, $\displaystyle \int_1^{\frac{n^2+2n+2}{n+2}}(-\ln x)\,dx=-\dfrac{n^2+2n+2}{n+2}\left(\ln\dfrac{n^2+2n+2}{n+2}-1\right)-1$이다.$\displaystyle b_n=-\dfrac4{(n+1)\ln(n+1)}\left\{\dfrac{n^2+2n+2}{n+2}\left(\ln\dfrac{n^2+2n+2}{n+2}-1\right)+1\right\}\quad(n\ge1),$$\displaystyle c_{n-1}=-\dfrac4{n\ln n}\left\{\dfrac{n^2+2n+2}{n+2}\left(\ln\dfrac{n^2+2n+2}{n+2}-1\right)+1\right\}\quad(n\ge2)$이라 하면 모든 자연수 $\displaystyle n$에 대하여 $\displaystyle c_n<a_n<b_n$이 성립한다.

이때, $\displaystyle \begin{aligned}\lim_{n\to\infty}b_n&=(-4)\lim_{n\to\infty}\left\{\dfrac{n^2+2n+2}{(n+1)(n+2)}\left(\dfrac{\ln\dfrac{n^2+2n+2}{(n+2)(n+1)}(n+1)}{\ln(n+1)}-\dfrac1{\ln(n+1)}\right)+\dfrac1{(n+1)\ln(n+1)}\right\}\\&=(-4)\lim_{n\to\infty}\left\{\dfrac{n^2+2n+2}{(n+1)(n+2)}\left(\dfrac{\ln\dfrac{n^2+2n+2}{(n+2)(n+1)}}{\ln(n+1)}+1-\dfrac1{\ln(n+1)}\right)+\dfrac1{(n+1)\ln(n+1)}\right\}\\&=-4\end{aligned}$이고$\displaystyle \begin{aligned}\lim_{n\to\infty}c_n&=(-4)\lim_{n\to\infty}\left\{\dfrac{(n+1)^2+2(n+1)+2}{(n+1)(n+3)}\left(\dfrac{\ln\dfrac{(n+1)^2+2(n+1)+2}{(n+3)(n+1)}(n+1)}{\ln(n+1)}-\dfrac1{\ln(n+1)}\right)+\dfrac1{(n+1)\ln(n+1)}\right\}\\&=(-4)\lim_{n\to\infty}\left\{\dfrac{(n+1)^2+2(n+1)+2}{(n+1)(n+3)}\left(\dfrac{\ln\dfrac{(n+1)^2+2(n+1)+2}{(n+3)(n+1)}}{\ln(n+1)}+1-\dfrac1{\ln(n+1)}\right)+\dfrac1{(n+1)\ln(n+1)}\right\}\\&=-4\end{aligned}$이므로 수열의 극한의 대소관계에 의해 $\displaystyle \lim_{n\to\infty}a_n=-4$이다.