(1) $\lim_{n \to \infty} \sum_{k=n+1}^{2n} \frac{n}{k^2 + 3kn + 2n^2}$ を求めよ。 (2) $\lim_{n \to \infty} n \sum_{k=1}^{2n} \frac{1}{(n+2k)^2}$ を求めよ。

1. 問題の内容

(1) $\lim_{n \to \infty} \sum_{k=n+1}^{2n} \frac{n}{k^2 + 3kn + 2n^2}$ を求めよ。

(2) $\lim_{n \to \infty} n \sum_{k=1}^{2n} \frac{1}{(n+2k)^2}$ を求めよ。

2. 解き方の手順

(1)

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

ここで，$k+n = nx+n, k+2n = nx+2n$ とすると，$k = nx$とできる.

$k=n+1$ のとき，$nx=n+1$だから，$x=1+\frac{1}{n}$.

$k=2n$ のとき，$nx=2n$だから，$x=2$.

よって，$k$を$nx$に置き換えることを考えると,

$\lim_{n \to \infty} \sum_{k=n+1}^{2n} \frac{n}{k^2 + 3kn + 2n^2} = \lim_{n \to \infty} \sum_{k=n+1}^{2n} \frac{n}{(k+n)(k+2n)} = \lim_{n \to \infty} \sum_{k=n+1}^{2n} \frac{n}{(k+n)(k+2n)} = \lim_{n \to \infty} \sum_{k=n+1}^{2n} \frac{n}{n^2} \frac{1}{(x+1)(x+2)} = \lim_{n \to \infty} \sum_{k=n+1}^{2n} \frac{1}{n} \frac{1}{(1+\frac{k}{n})(2+\frac{k}{n})}$

ここで，$f(x) = \frac{1}{(x+n)(x+2n)} = \frac{1}{x+n} - \frac{1}{x+2n}$ とすると、部分分数分解をすると

$\frac{1}{k+n} - \frac{1}{k+2n}$

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

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

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

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

$= 2\sum_{k=1}^{3n} \frac{1}{k} - \sum_{k=1}^{2n} \frac{1}{k} - \sum_{k=1}^{4n} \frac{1}{k}$

$\lim_{n \to \infty} \sum_{k=n+1}^{2n} \frac{n}{k^2+3kn+2n^2} = \int_1^2 \frac{1}{(x+1)(x+2)}dx = \int_1^2 (\frac{1}{x+1} - \frac{1}{x+2})dx$

$= [\ln(x+1) - \ln(x+2)]_1^2 = [\ln\frac{x+1}{x+2}]_1^2 = \ln\frac{3}{4} - \ln\frac{2}{3} = \ln\frac{3}{4} \times \frac{3}{2} = \ln\frac{9}{8}$

(2)

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

$= \int_0^2 \frac{1}{(1+2x)^2}dx$

$u=1+2x$ と置くと、$du = 2dx$

$\int_1^5 \frac{1}{u^2}\frac{du}{2} = \frac{1}{2} \int_1^5 u^{-2}du = \frac{1}{2} [-u^{-1}]_1^5 = \frac{1}{2}[-\frac{1}{u}]_1^5 = \frac{1}{2}[-\frac{1}{5} - (-1)] = \frac{1}{2} [1-\frac{1}{5}] = \frac{1}{2}[\frac{4}{5}] = \frac{2}{5}$

3. 最終的な答え

(1) $\ln\frac{9}{8}$

(2) $\frac{2}{5}$