자연계열 (오전) 1번

1. $\displaystyle A_n=\pi\int_0^{2^na_n}(2^na_n-y)dy=\left.\pi\left(2^na_ny-\frac12y^2\right)\right|_0^{2^na_n}=\frac\pi2 4^na_n^2$이다.

$\displaystyle A_n=2\pi$이므로 $\displaystyle a_n=\frac2{2^n}$이고 $\displaystyle \sum_{n=1}^\infty a_n=2$

2. $\displaystyle B_n=\int_0^1\left(1-\frac1{2^nb_n}\ln(x+1)\right)dx=\left.x-\frac1{2^nb_n}((x+1)\ln(x+1)-(x+1))\right|_0^1=1-\frac1{2^nb_n}(2\ln2-1)$이다.

$\displaystyle B_n=2\pi$이므로 $\displaystyle b_n=\frac{2\ln2-1}{2^n(1-2\pi)}$이고 $\displaystyle \sum_{n=1}^\infty b_n=\frac{2\ln2-1}{1-2\pi}=\frac{1-2\ln2}{2\pi-1}=-\frac{2\ln2-1}{2\pi-1}$

3. $\displaystyle C_n=\pi\int_0^1c_n^24^{-nx}dx=\left.\frac{\pi c_n^2}{\ln4^{-n}}4^{-nx}\right|_0^1=\frac\pi{2\ln2}\left(1-\frac1{4^n}\right)\frac{c_n^2}n$이다.$\displaystyle C_n=2\pi$이므로 $\displaystyle c_n=\sqrt{4\ln2\cdot n}\cdot\frac{2^n}{\sqrt{4^n-1}}$이고 $\displaystyle \frac{c_n}{\sqrt n}=\sqrt{4\ln2}\cdot\frac{2^n}{\sqrt{4^n-1}}$이다.

따라서 $\displaystyle \lim_{n\to\infty}c_n=\lim_{n\to\infty}\sqrt{4\ln2\cdot n}\cdot\frac{2^n}{\sqrt{4^n-1}}=\infty$이고

$\displaystyle \lim_{n\to\infty}\frac{c_n}{\sqrt n}=\lim_{n\to\infty}\sqrt{4\ln2}\cdot\frac{2^n}{\sqrt{4^n-1}}=\sqrt{4\ln2}=2\sqrt{\ln2}\ne0$이다.

“무한급수 $\displaystyle \sum_{n=1}^\infty d_n$ 수렴하면, $\displaystyle \lim_{n\to\infty}d_n=0$이다”는 사실에 의하여무한급수 $\displaystyle \sum_{n=1}^\infty c_n$와 $\displaystyle \sum_{n=1}^\infty\frac{c_n}{\sqrt n}$는 모두 발산한다.