자연계열 2번

그래프는 다음과 같다.

$\displaystyle a_n=\int_{1/2^n}^{1/2^{n-1}}x\,dx,\qquad b_n=\int_{1/2^n}^{1/2^{n-1}}2x\,dx$라고 하면$\displaystyle a_n=\left.\frac12x^2\right|_{1/2^n}^{1/2^{n-1}}=\frac12\left(\frac1{2^{2n-2}}-\frac1{2^{2n}}\right)=\frac3{2^{2n+1}},$$\displaystyle b_n=\left.x^2\right|_{1/2^n}^{1/2^{n-1}}=\frac1{2^{2n-2}}-\frac1{2^{2n}}=\frac3{2^{2n}}$이 된다.

$\displaystyle \begin{aligned}a_1+a_3+\cdots+a_{2m+1}&=3\left(\frac1{2^3}+\frac1{2^7}+\cdots+\frac1{2^{4m+3}}\right)\\&=\frac25\left(1-\frac1{2^{4(m+1)}}\right)\end{aligned}$이고$\displaystyle \begin{aligned}b_2+b_4+\cdots+b_{2m}&=3\left(\frac1{2^4}+\frac1{2^8}+\cdots+\frac1{2^{4m}}\right)\\&=\frac15\left(1-\frac1{2^{4m}}\right)\end{aligned}$이므로

(1) $\displaystyle n=2m+1$일 때$\displaystyle \begin{aligned}S_n&=\frac25\left(1-\frac1{2^{4(m+1)}}\right)+\frac15\left(1-\frac1{2^{4m}}\right)\\&=\frac15\left(3-\frac1{2^{4m}}-\frac1{2^{4m+3}}\right)\end{aligned}$(2) $\displaystyle n=2m$일 때$\displaystyle \begin{aligned}S_n&=\frac25\left(1-\frac1{2^{4m}}\right)+\frac15\left(1-\frac1{2^{4m}}\right)\\&=\frac15\left(3-\frac1{2^{4m-1}}-\frac1{2^{4m}}\right)\end{aligned}$이 된다.

따라서 $\displaystyle n$이 무한대로 갈 때 $\displaystyle S_n$은 $\displaystyle \frac35$으로 수렴한다.