## 1. 問題の内容

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1. 問題の内容

以下の4つの数列の和を $n$ の式で表し、さらに $n \to \infty$ における極限を求めます。

(1) $\frac{1 \cdot n}{n} + \frac{2 \cdot (n-1)}{n} + \frac{3 \cdot (n-2)}{n} + \dots + \frac{n \cdot 1}{n}$

(2) $\frac{1}{1 \cdot 3 \cdot 5} + \frac{1}{3 \cdot 5 \cdot 7} + \frac{1}{5 \cdot 7 \cdot 9} + \dots + \frac{1}{(2n-1)(2n+1)(2n+3)}$

(3) $1 - \frac{2}{3} + \frac{3}{9} - \frac{4}{27} + \dots + \frac{n}{(-3)^{n-1}}$

(4) $\frac{3}{1^2 \cdot 2^2} + \frac{5}{2^2 \cdot 3^2} + \frac{7}{3^2 \cdot 4^2} + \dots + \frac{2n+1}{n^2(n+1)^2}$

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2. 解き方の手順

### (1)

まず、一般項を求めます。第 $k$ 項は $\frac{k(n - k + 1)}{n}$ と表せます。

したがって、和 $S_n$ は次のようになります。

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

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

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

$S_n = \frac{1}{n} \left[ n \frac{n(n+1)}{2} - \frac{n(n+1)(2n+1)}{6} + \frac{n(n+1)}{2} \right]$

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

$S_n = \frac{(n+1)(n+2)}{6}$

次に、$n \to \infty$ のときの極限を求めます。

$\lim_{n \to \infty} \frac{(n+1)(n+2)}{6} = \lim_{n \to \infty} \frac{n^2 + 3n + 2}{6} = \infty$

### (2)

一般項を部分分数分解します。

$\frac{1}{(2k-1)(2k+1)(2k+3)} = \frac{A}{2k-1} + \frac{B}{2k+1} + \frac{C}{2k+3}$

$1 = A(2k+1)(2k+3) + B(2k-1)(2k+3) + C(2k-1)(2k+1)$

$k = \frac{1}{2}$ のとき、$1 = A(2)(4) = 8A \implies A = \frac{1}{8}$

$k = -\frac{1}{2}$ のとき、$1 = B(-2)(2) = -4B \implies B = -\frac{1}{4}$

$k = -\frac{3}{2}$ のとき、$1 = C(-4)(-2) = 8C \implies C = \frac{1}{8}$

$\frac{1}{(2k-1)(2k+1)(2k+3)} = \frac{1}{8} \left( \frac{1}{2k-1} - \frac{2}{2k+1} + \frac{1}{2k+3} \right) = \frac{1}{8} \left[ (\frac{1}{2k-1} - \frac{1}{2k+1}) - (\frac{1}{2k+1} - \frac{1}{2k+3}) \right]$

$S_n = \sum_{k=1}^{n} \frac{1}{(2k-1)(2k+1)(2k+3)} = \frac{1}{8} \sum_{k=1}^{n} \left[ (\frac{1}{2k-1} - \frac{1}{2k+1}) - (\frac{1}{2k+1} - \frac{1}{2k+3}) \right]$

$S_n = \frac{1}{8} \left[ (1 - \frac{1}{3}) + (\frac{1}{3} - \frac{1}{5}) + \dots + (\frac{1}{2n-1} - \frac{1}{2n+1}) - (\frac{1}{3} - \frac{1}{5}) - (\frac{1}{5} - \frac{1}{7}) - \dots - (\frac{1}{2n+1} - \frac{1}{2n+3}) \right]$

$S_n = \frac{1}{8} \left[ 1 - \frac{1}{3} - \frac{1}{2n+1} + \frac{1}{2n+3} \right] = \frac{1}{8} \left( \frac{2}{3} - \frac{2}{(2n+1)(2n+3)} \right) = \frac{1}{4} \left( \frac{1}{3} - \frac{1}{(2n+1)(2n+3)} \right)$

$S_n = \frac{1}{4} \left( \frac{(2n+1)(2n+3) - 3}{3(2n+1)(2n+3)} \right) = \frac{1}{4} \frac{4n^2 + 8n}{3(2n+1)(2n+3)} = \frac{n(n+2)}{3(2n+1)(2n+3)}$

$\lim_{n \to \infty} \frac{n(n+2)}{3(2n+1)(2n+3)} = \lim_{n \to \infty} \frac{n^2+2n}{12n^2 + 18n + 9} = \frac{1}{12}$

### (3)

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

これは等比数列と等差数列の積の和です。

$S_n = 1 - \frac{2}{3} + \frac{3}{9} - \frac{4}{27} + \dots + \frac{n}{(-3)^{n-1}}$

$-\frac{1}{3} S_n = -\frac{1}{3} + \frac{2}{9} - \frac{3}{27} + \dots - \frac{n-1}{(-3)^{n-1}} - \frac{n}{(-3)^n}$

$\frac{4}{3} S_n = 1 - \frac{1}{3} + \frac{1}{9} - \frac{1}{27} + \dots + \frac{1}{(-3)^{n-1}} - \frac{n}{(-3)^n}$

$\frac{4}{3} S_n = \sum_{k=0}^{n-1} \frac{1}{(-3)^k} - \frac{n}{(-3)^n}$

$\sum_{k=0}^{n-1} \frac{1}{(-3)^k} = \frac{1 - (-\frac{1}{3})^n}{1 - (-\frac{1}{3})} = \frac{1 - (-\frac{1}{3})^n}{\frac{4}{3}} = \frac{3}{4} (1 - (-\frac{1}{3})^n)$

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

$S_n = \frac{9}{16} (1 - (-\frac{1}{3})^n) - \frac{3}{4} \frac{n}{(-3)^n} = \frac{9}{16} - \frac{9}{16} (-\frac{1}{3})^n - \frac{3n}{4(-3)^n} = \frac{9}{16} + \frac{9}{16} \frac{(-1)^{n+1}}{3^n} + \frac{3n(-1)^{n+1}}{4 \cdot 3^n}$

$\lim_{n \to \infty} S_n = \lim_{n \to \infty} \left[ \frac{9}{16} + \frac{9}{16} \frac{(-1)^{n+1}}{3^n} + \frac{3n(-1)^{n+1}}{4 \cdot 3^n} \right]$

$\lim_{n \to \infty} S_n = \frac{9}{16} + 0 + 0 = \frac{9}{16}$

### (4)

$\frac{2k+1}{k^2(k+1)^2} = \frac{(k+1)^2 - k^2}{k^2(k+1)^2} = \frac{1}{k^2} - \frac{1}{(k+1)^2}$

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

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

$\lim_{n \to \infty} S_n = \lim_{n \to \infty} \left( 1 - \frac{1}{(n+1)^2} \right) = 1 - 0 = 1$

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3. 最終的な答え

(1) $S_n = \frac{(n+1)(n+2)}{6}$, $\lim_{n \to \infty} S_n = \infty$

(2) $S_n = \frac{n(n+2)}{3(2n+1)(2n+3)}$, $\lim_{n \to \infty} S_n = \frac{1}{12}$

(3) $S_n = \frac{9}{16} + \frac{9}{16} \frac{(-1)^{n+1}}{3^n} + \frac{3n(-1)^{n+1}}{4 \cdot 3^n}$, $\lim_{n \to \infty} S_n = \frac{9}{16}$

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