자연계열 2번

어떤 자연수 $\displaystyle a$ ($\displaystyle 1\le a\le n$)에 대하여 $\displaystyle n$번의 시행 후 자연수 $\displaystyle a$가 보이게 뒤집어져 있을 확률을 $\displaystyle \mathrm{P}(a)$라 하면$\displaystyle \mathrm{P}(a)={}_{n}\mathrm{C}_{1}\!\left(\frac{1}{n}\right)^{1}\!\left(1-\frac{1}{n}\right)^{n-1}+{}_{n}\mathrm{C}_{3}\!\left(\frac{1}{n}\right)^{3}\!\left(1-\frac{1}{n}\right)^{n-3}+{}_{n}\mathrm{C}_{5}\!\left(\frac{1}{n}\right)^{5}\!\left(1-\frac{1}{n}\right)^{n-5}+\cdots+{}_{n}\mathrm{C}_{m}\!\left(\frac{1}{n}\right)^{m}\!\left(1-\frac{1}{n}\right)^{n-m}$(단, $\displaystyle m$은 $\displaystyle n$ 이하의 자연수 중 최대인 홀수)이고,이항정리에 의하여$\displaystyle \mathrm{P}(a)=\frac{\displaystyle\sum_{k=0}^{n}{}_{n}\mathrm{C}_{k}\!\left(\frac{1}{n}\right)^{k}\!\left(1-\frac{1}{n}\right)^{n-k}-\sum_{k=0}^{n}(-1)^{k}{}_{n}\mathrm{C}_{k}\!\left(\frac{1}{n}\right)^{k}\!\left(1-\frac{1}{n}\right)^{n-k}}{2}$

한편, $\displaystyle \sum_{k=0}^{n}{}_{n}\mathrm{C}_{k}\!\left(\frac{1}{n}\right)^{k}\!\left(1-\frac{1}{n}\right)^{n-k}=\left(\frac{1}{n}+1-\frac{1}{n}\right)^{n}=1$ 이고$\displaystyle \sum_{k=0}^{n}(-1)^{k}{}_{n}\mathrm{C}_{k}\!\left(\frac{1}{n}\right)^{k}\!\left(1-\frac{1}{n}\right)^{n-k}=\sum_{k=0}^{n}{}_{n}\mathrm{C}_{k}\!\left(-\frac{1}{n}\right)^{k}\!\left(1-\frac{1}{n}\right)^{n-k}=\left(-\frac{1}{n}+1-\frac{1}{n}\right)^{n}=\left(1-\frac{2}{n}\right)^{n}$이므로$\displaystyle \mathrm{P}(a)=\frac{1-\left(1-\dfrac{2}{n}\right)^{n}}{2}$

따라서$\displaystyle a_{n}=\sum_{a=1}^{n}\left\{a\times\frac{1-\left(1-\frac{2}{n}\right)^{n}}{2}\right\}=\frac{n(n+1)}{2}\times\frac{1-\left(1-\frac{2}{n}\right)^{n}}{2}$

이므로$\displaystyle \lim_{n\to\infty}\frac{a_{n}}{n^{2}}=\lim_{n\to\infty}\left\{\frac{n(n+1)}{2n^{2}}\times\frac{1-\left(1-\frac{2}{n}\right)^{n}}{2}\right\}=\frac{e^{2}-1}{4e^{2}}$