자연계열 4번

먼저 자연수 $\displaystyle k=1,\,2,\,\cdots,\,3n$에 대해 적분값$\displaystyle \int_{0}^{k\pi}\left|\sin\frac{x}{2n}\right|dx$을 구하자.

(i) $\displaystyle k=1,\,2,\,\cdots,\,2n$인 경우$\displaystyle 0\le x\le2n\pi$에서 $\displaystyle \sin\dfrac{x}{2n}\ge0$이므로$\displaystyle \int_{0}^{k\pi}\left|\sin\frac{x}{2n}\right|dx=\int_{0}^{k\pi}\sin\frac{x}{2n}\,dx=\Biggl[-2n\cos\frac{x}{2n}\Biggr]_{0}^{k\pi}=2n\left(1-\cos\frac{k\pi}{2n}\right)$이다.

(ii) $\displaystyle k=2n+1,\,2n+2,\,\cdots,\,3n$인 경우$\displaystyle 0\le x<2n\pi$에서 $\displaystyle \sin\dfrac{x}{2n}\ge0$이고, $\displaystyle 2n\pi\le x\le3n\pi$에서 $\displaystyle \sin\dfrac{x}{2n}\le0$이므로$\displaystyle \int_{0}^{k\pi}\left|\sin\frac{x}{2n}\right|dx=\int_{0}^{2n\pi}\sin\frac{x}{2n}\,dx+\int_{2n\pi}^{k\pi}\left(-\sin\frac{x}{2n}\right)dx=\Biggl[-2n\cos\frac{x}{2n}\Biggr]_{0}^{2n\pi}+\Biggl[2n\cos\frac{x}{2n}\Biggr]_{2n\pi}^{k\pi}=2n\left(3+\cos\frac{k\pi}{2n}\right)$이다.

위 (i)과 (ii)에 의해$\displaystyle \begin{aligned} T_{n} &=\sum_{k=1}^{n}\frac{1}{n^{2}}\left\{\int_{0}^{k\pi}\left|\sin\frac{x}{2n}\right|dx+\int_{0}^{(n+k)\pi}\left|\sin\frac{x}{2n}\right|dx+\int_{0}^{(2n+k)\pi}\left|\sin\frac{x}{2n}\right|dx\right\} \\ &=\sum_{k=1}^{n}\frac{1}{n}\left[2\left(1-\cos\frac{k\pi}{2n}\right)+2\left\{1-\cos\frac{(n+k)\pi}{2n}\right\}+2\left\{3+\cos\frac{(2n+k)\pi}{2n}\right\}\right] \\ &=10+\frac{1}{n}\sum_{k=1}^{n}\left(-4\cos\frac{k\pi}{2n}+2\sin\frac{k\pi}{2n}\right) \end{aligned}$이다.

따라서 정적분과 급수의 합 사이의 관계로부터$\displaystyle \begin{aligned} \lim_{n\to\infty}T_{n} &=10+\lim_{n\to\infty}\frac{1}{n}\sum_{k=1}^{n}\left(-4\cos\frac{k\pi}{2n}+2\sin\frac{k\pi}{2n}\right) \\ &=10+\int_{0}^{1}\left(-4\cos\frac{\pi}{2}x+2\sin\frac{\pi}{2}x\right)dx \\ &=10+\Biggl[-\frac{8}{\pi}\sin\frac{\pi}{2}x-\frac{4}{\pi}\cos\frac{\pi}{2}x\Biggr]_{0}^{1}=10-\frac{4}{\pi} \end{aligned}$이다.