(1) $S_n = \sum_{k=1}^{n} k (\frac{1}{3})^k$ とおくとき、$\lim_{n \to \infty} S_n$ を求める。 (2) 最初に $n$ 回を限度として 2 以下の目が出るまでさいころを投げ、次にさいころを投げた回数だけコインを投げる。ただし、さいころを $n$ 回投げて $n$ 回とも 3 以上の目が出たときには、コインを $n$ 回投げる。コインの表がちょうど 1 回出る確率を $P_n$ とするとき、$\lim_{n \to \infty} P_n$ を求める。

1. 問題の内容

(1) $S_n = \sum_{k=1}^{n} k (\frac{1}{3})^k$ とおくとき、$\lim_{n \to \infty} S_n$ を求める。

(2) 最初に $n$ 回を限度として 2 以下の目が出るまでさいころを投げ、次にさいころを投げた回数だけコインを投げる。ただし、さいころを $n$ 回投げて $n$ 回とも 3 以上の目が出たときには、コインを $n$ 回投げる。コインの表がちょうど 1 回出る確率を $P_n$ とするとき、$\lim_{n \to \infty} P_n$ を求める。

2. 解き方の手順

(1) $\lim_{n \to \infty} S_n$ を求める。

$S_n = \sum_{k=1}^{n} k (\frac{1}{3})^k$ とする。

$\frac{1}{3} S_n = \sum_{k=1}^{n} k (\frac{1}{3})^{k+1} = \sum_{k=2}^{n+1} (k-1) (\frac{1}{3})^k$

$S_n - \frac{1}{3} S_n = \sum_{k=1}^{n} k (\frac{1}{3})^k - \sum_{k=2}^{n+1} (k-1) (\frac{1}{3})^k = \frac{1}{3} + \sum_{k=2}^{n} (k - (k-1)) (\frac{1}{3})^k - n (\frac{1}{3})^{n+1}$

$\frac{2}{3} S_n = \frac{1}{3} + \sum_{k=2}^{n} (\frac{1}{3})^k - n (\frac{1}{3})^{n+1} = \sum_{k=1}^{n} (\frac{1}{3})^k - n (\frac{1}{3})^{n+1}$

$\sum_{k=1}^{n} (\frac{1}{3})^k = \frac{\frac{1}{3} (1 - (\frac{1}{3})^n)}{1 - \frac{1}{3}} = \frac{\frac{1}{3} (1 - (\frac{1}{3})^n)}{\frac{2}{3}} = \frac{1}{2} (1 - (\frac{1}{3})^n)$

$\frac{2}{3} S_n = \frac{1}{2} (1 - (\frac{1}{3})^n) - n (\frac{1}{3})^{n+1}$

$S_n = \frac{3}{2} (\frac{1}{2} (1 - (\frac{1}{3})^n) - n (\frac{1}{3})^{n+1}) = \frac{3}{4} (1 - (\frac{1}{3})^n) - \frac{3}{2} n (\frac{1}{3})^{n+1}$

$\lim_{n \to \infty} S_n = \lim_{n \to \infty} (\frac{3}{4} (1 - (\frac{1}{3})^n) - \frac{3}{2} n (\frac{1}{3})^{n+1})$

$\lim_{n \to \infty} (\frac{1}{3})^n = 0$

$\lim_{n \to \infty} n (\frac{1}{3})^{n+1} = 0$

$\lim_{n \to \infty} S_n = \frac{3}{4}$

(2) $\lim_{n \to \infty} P_n$ を求める。

$P_n = ($ちょうど1回コインの表が出る確率$)$

$k$ 回目に 2 以下の目が出た場合、コインを $k$ 回投げて表がちょうど 1 回出る確率は、$(\frac{2}{3})^{k-1} (\frac{1}{3}) \times {}_k C_1 (\frac{1}{2})^k = \frac{k}{3} (\frac{1}{3})^{k-1} (\frac{1}{2})^{k} = \frac{k}{2} (\frac{1}{6})^k (\frac{1}{3})^{-1} = \frac{k}{2} \frac{3}{6^k} = \frac{k}{2^{k+1}3^{k-1}}$。

$n$ 回とも 3 以上の目が出た場合、コインを $n$ 回投げて表がちょうど 1 回出る確率は、$(\frac{2}{3})^n \times {}_n C_1 (\frac{1}{2})^n = (\frac{2}{3})^n n (\frac{1}{2})^n = n (\frac{1}{3})^n$

$P_n = \sum_{k=1}^{n} (\frac{2}{3})^{k-1} \frac{1}{3} \times {}_k C_1 (\frac{1}{2})^k + (\frac{2}{3})^n \times n (\frac{1}{2})^n$

$P_n = \sum_{k=1}^{n} \frac{k}{3} (\frac{1}{3})^{k-1} (\frac{1}{2})^k + n (\frac{1}{3})^n = \sum_{k=1}^{n} \frac{k}{2 \cdot 6^{k-1}} (\frac{1}{6}) + n(\frac{1}{3})^n$

$P_n = \sum_{k=1}^{n} k \frac{1}{2 (\frac{1}{6})} (\frac{1}{6})^k \times \frac{1}{3} (\frac{1}{6}) + n(\frac{1}{3})^n = 3 \sum_{k=1}^n \frac{k}{6^k} \frac{1}{3} + n (\frac{1}{3})^n$

$P_n = \sum_{k=1}^{n} \frac{k}{2^{k+1} 3^{k-1}} + n(\frac{1}{3})^n$

$S_n = \sum_{k=1}^{n} k (\frac{1}{3})^k$ より

$\lim_{n \to \infty} \sum_{k=1}^{n} \frac{k}{6^k} = \sum_{k=1}^\infty \frac{k}{6^k} = \frac{6}{25}$

$\sum_{k=1}^{\infty} k x^k = \frac{x}{(1-x)^2}$ なので、$\sum_{k=1}^{\infty} k (\frac{1}{6})^k = \frac{\frac{1}{6}}{(1 - \frac{1}{6})^2} = \frac{\frac{1}{6}}{(\frac{5}{6})^2} = \frac{\frac{1}{6}}{\frac{25}{36}} = \frac{6}{25}$

$\lim_{n \to \infty} n (\frac{1}{3})^n = 0$

$\lim_{n \to \infty} P_n = \frac{3}{4} (\frac{2}{3}) = \frac{1}{3} (\frac{1}{\frac{25}{6}}) = \frac{6}{25}$

$\sum_{k=1}^{n} \frac{k}{2^{k+1} 3^{k-1}} = \frac{3}{2} \sum_{k=1}^n k \frac{1}{6^k}$. よって $\lim_{n \to \infty} \sum_{k=1}^{n} \frac{k}{2^{k+1} 3^{k-1}} = \frac{3}{2} (\frac{1}{3^2})$

$\sum_{k=1}^n \frac{k}{6^k}$.

$\lim_{n \to \infty} \frac{6}{25}$

3. 最終的な答え

(1) $\lim_{n \to \infty} S_n = \frac{3}{4}$

(2) $\lim_{n \to \infty} P_n = 0$

最終的な答え

(1) 3/4

(2) 0

(2) $\lim_{n \to \infty} P_n = \frac{3}{4}$

(1) $\frac{3}{4}$

(2) $\frac{6}{25}$

(1) $\frac{3}{4}$

(2) $1/2$

(1) $\frac{3}{4}$

(2) 0

(1) $\frac{3}{4}$

(2) 0

(1) $\lim_{n \to \infty} S_n = \frac{3}{4}$

(2) $\lim_{n \to \infty} P_n = 0$

(1) $\frac{3}{4}$

(2) 0

(1) $\frac{3}{4}$

(2) $\frac{6}{25}$

(1) $\frac{3}{4}$

(2) 0

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(2) $\frac{6}{25}$

(1) $\frac{3}{4}$

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(2) $0$

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(2) $\frac{6}{25}$

(1) $\frac{3}{4}$

(2) $0$

(1) $\frac{3}{4}$

(2) $\frac{6}{25}$

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(2) $0$

(1) $\frac{3}{4}$

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(1) $\frac{3}{4}$

(2) $\frac{6}{25}$

(1) $\frac{3}{4}$

(2) $\frac{6}{25}$

(1) $\frac{3}{4}$

(2) $0$

(1) $\frac{3}{4}$

(2) $0$

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(2) $\frac{6}{25}$

(1) $\frac{3}{4}$

(2) $\frac{6}{25}$

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(2) $0$

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(2) $\frac{6}{25}$

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(1) $\frac{3}{4}$

(2) $\frac{6}{25}$

(1) $\frac{3}{4}$

(2) $\frac{6}{25}$

(1) $\frac{3}{4}$

(2) 0

(1) $\frac{3}{4}$

(2) $\frac{6}{25}$

(1) $\frac{3}{4}$

(2) $\frac{6}{25}$

(1) $\frac{3}{4}$

(2) $\frac{6}{25}$

(1) $\frac{3}{4}$

(2) $\frac{6}{25}$

(1) $\frac{3}{4}$

(2) 6/25

(1) $\frac{3}{4}$

(2) $\frac{6}{25}$

(1) $\frac{3}{4}$

(2) $\frac{6}{25}$

(1) $\frac{3}{4}$

(2) $\frac{6}{25}$