統計検定 1級 過去問 解答/解答例と解説
2015年11月29日 (日) 試験

$$\begin{equation} f(k) = P(X = k) = \frac{1}{k(k+1)} > 0 \end{equation}$$

$$\begin{equation} f(k) = \frac{1}{k(k+1)} = \frac{a}{k} + \frac{b}{k+1} \end{equation}$$

$$\begin{align} \frac{1}{k(k+1)} &= \frac{a}{k} + \frac{b}{k+1} \\ &= \frac{a(k+1) + bk}{k(k+1)} \\ &= \frac{ak+a + bk}{k(k+1)} \\ &= \frac{(a + b)k + a}{k(k+1)} \\ \end{align}$$

$$\begin{equation} \begin{cases} a + b = 0 \\ a = 1 \end{cases} \end{equation}$$

$$\begin{equation} a = 1,~b = -1 \end{equation}$$

$$\begin{equation} f(k) = \frac{1}{k(k+1)} = \frac{1}{k} - \frac{1}{k+1} \end{equation}$$

$$\begin{align} \sum_{k=1}^{\infty}f(k) &= \sum_{k=1}^{\infty}\left(\frac{1}{k} - \frac{1}{k+1}\right) \\ &= \lim_{n \to \infty}\sum_{k=1}^{n}\left(\frac{1}{k} - \frac{1}{k+1}\right) \\ &= \lim_{n \to \infty}\left(\frac{1}{1} - \frac{1}{2} + \frac{1}{2} - \frac{1}{3} + \ldots + \frac{1}{n} - \frac{1}{n+1} \right) \\ &= \lim_{n \to \infty}\left(1 - \frac{1}{n+1} \right) \\ &= 1 \end{align}$$