## 問題の解答

## 問題の解答

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1. 問題の内容

与えられた数列の極限を求める問題です。問題は以下の6つの数列について極限を求めるものです。

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

(2) $a_n = n(\sqrt{1+\frac{1}{n}} - 1)$

(3) $a_n = \frac{1! + 2! + \dots + n!}{n!}$

(4) $a_n = \frac{1}{\log n} + \frac{1}{\log (n^2)} + \dots + \frac{1}{\log (n^n)}$

(5) $a_n = \frac{2^n}{n!}$

(6) $a_n = \frac{n^3}{2^n}$

以下、それぞれの数列について極限を求めます。

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2. 解き方の手順

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

まず、分子の和を計算します。

$1^2 + 3^2 + \dots + (2n-1)^2 = \sum_{k=1}^{n} (2k-1)^2 = \sum_{k=1}^{n} (4k^2 - 4k + 1) = 4 \sum_{k=1}^{n} k^2 - 4 \sum_{k=1}^{n} k + \sum_{k=1}^{n} 1$

$= 4 \frac{n(n+1)(2n+1)}{6} - 4 \frac{n(n+1)}{2} + n = \frac{2n(n+1)(2n+1)}{3} - 2n(n+1) + n$

$= \frac{n}{3}(2(n+1)(2n+1) - 6(n+1) + 3) = \frac{n}{3}(4n^2 + 6n + 2 - 6n - 6 + 3) = \frac{n}{3}(4n^2 - 1) = \frac{4n^3 - n}{3}$

したがって、$a_n = \frac{4n^3 - n}{3n^3} = \frac{4}{3} - \frac{1}{3n^2}$

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

(2) $a_n = n(\sqrt{1+\frac{1}{n}} - 1)$

$\sqrt{1+\frac{1}{n}} = 1 + \frac{1}{2} \cdot \frac{1}{n} + O(\frac{1}{n^2})$（テイラー展開）

$a_n = n(1 + \frac{1}{2n} + O(\frac{1}{n^2}) - 1) = n(\frac{1}{2n} + O(\frac{1}{n^2})) = \frac{1}{2} + O(\frac{1}{n})$

$\lim_{n \to \infty} a_n = \lim_{n \to \infty} (\frac{1}{2} + O(\frac{1}{n})) = \frac{1}{2}$

または、

$a_n = n(\sqrt{1+\frac{1}{n}} - 1) = n \frac{(1+\frac{1}{n}) - 1}{\sqrt{1+\frac{1}{n}} + 1} = \frac{n \cdot \frac{1}{n}}{\sqrt{1+\frac{1}{n}} + 1} = \frac{1}{\sqrt{1+\frac{1}{n}} + 1}$

$\lim_{n \to \infty} a_n = \frac{1}{\sqrt{1+0} + 1} = \frac{1}{2}$

(3) $a_n = \frac{1! + 2! + \dots + n!}{n!}$

$a_n = \frac{1!}{n!} + \frac{2!}{n!} + \dots + \frac{(n-1)!}{n!} + \frac{n!}{n!} = \frac{1}{n!} + \frac{2!}{n!} + \dots + \frac{1}{n} + 1$

$a_n = 1 + \frac{1}{n} + \frac{1}{n(n-1)} + \dots + \frac{1}{n!}$

$a_n < 1 + \frac{1}{n} + \frac{1}{n^2} + \dots + \frac{1}{n^{n-1}} < \frac{1}{1 - \frac{1}{n}} = \frac{n}{n-1}$

$1 < a_n < \frac{n}{n-1}$

$\lim_{n \to \infty} \frac{n}{n-1} = 1$ なので、挟みうちの原理より $\lim_{n \to \infty} a_n = 1$

(4) $a_n = \frac{1}{\log n} + \frac{1}{\log (n^2)} + \dots + \frac{1}{\log (n^n)}$

$a_n = \frac{1}{\log n} + \frac{1}{2 \log n} + \dots + \frac{1}{n \log n} = \frac{1}{\log n} (1 + \frac{1}{2} + \dots + \frac{1}{n})$

$a_n = \frac{1}{\log n} \sum_{k=1}^{n} \frac{1}{k}$

調和級数 $\sum_{k=1}^{n} \frac{1}{k} \approx \log n + \gamma$ ($\gamma$はオイラー・マスケローニ定数)

$a_n \approx \frac{\log n + \gamma}{\log n} = 1 + \frac{\gamma}{\log n}$

$\lim_{n \to \infty} a_n = \lim_{n \to \infty} (1 + \frac{\gamma}{\log n}) = 1$

(5) $a_n = \frac{2^n}{n!}$

$a_{n+1} = \frac{2^{n+1}}{(n+1)!} = \frac{2 \cdot 2^n}{(n+1) n!} = \frac{2}{n+1} a_n$

$\frac{a_{n+1}}{a_n} = \frac{2}{n+1}$

$\lim_{n \to \infty} \frac{a_{n+1}}{a_n} = \lim_{n \to \infty} \frac{2}{n+1} = 0 < 1$

したがって、$\lim_{n \to \infty} a_n = 0$

(6) $a_n = \frac{n^3}{2^n}$

$a_{n+1} = \frac{(n+1)^3}{2^{n+1}} = \frac{(n+1)^3}{2 \cdot 2^n} = \frac{1}{2} \frac{(n+1)^3}{n^3} a_n$

$\frac{a_{n+1}}{a_n} = \frac{1}{2} (\frac{n+1}{n})^3 = \frac{1}{2} (1+\frac{1}{n})^3$

$\lim_{n \to \infty} \frac{a_{n+1}}{a_n} = \lim_{n \to \infty} \frac{1}{2} (1+\frac{1}{n})^3 = \frac{1}{2} < 1$

したがって、$\lim_{n \to \infty} a_n = 0$

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3. 最終的な答え

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

(2) $\frac{1}{2}$

(3) $1$

(4) $1$

(5) $0$

(6) $0$