自然数 \(n\) に対し \[\begin{align} S _ n & = \displaystyle\int _ 0^1 \dfrac{1 -(-x)^n}{1+x} \, dx \\ T _ n & = \textstyle\sum\limits _ {k=1}^n \dfrac{(-1)^{k-1}}{k( k+1 )} \end{align}\] とおく. このとき以下の各問いに答えよ.
- (1) 次の不等式を示せ. \[ \left| S _ n -\displaystyle\int _ 0^1 \dfrac{1}{1+x} \, dx \right| \leqq \dfrac{1}{n+1} \]
(2) \(T _ n -2S _ n\) を \(n\) を用いて表せ.
(3) 極限値 \(\displaystyle\lim _ {n \rightarrow \infty} T _ n\) を求めよ.
【 解 答 】
(1)
\(0 \leqq x \leqq 1\) において, \(\dfrac{x^n}{1+x} \leqq x^n\) であることを用いて \[\begin{align} \left| S _ n -\displaystyle\int _ 0^1 \dfrac{1}{1+x} \, dx \right| & = \left| \displaystyle\int _ 0^1 \dfrac{(-x)^n}{1+x} \, dx \right| \\ & = \displaystyle\int _ 0^1 \dfrac{n^n}{1+x} \, dx \leqq \displaystyle\int _ 0^1 x^n \, dx \\ & = \left[ \dfrac{x^{n+1}}{n+1} \right] _ 0^1 = \dfrac{1}{n+1} \end{align}\] ゆえに \[ \left| S _ n -\displaystyle\int _ 0^1 \dfrac{1}{1+x} \, dx \right| \leqq \dfrac{1}{n+1} \]
(2)
\[\begin{align} S _ n & = \displaystyle\int _ 0^1 \dfrac{1-(-x)^n}{1-(-x)} \, dx = \textstyle\sum\limits _ {k=1}^n \displaystyle\int _ 0^1 (-x)^{k-1} \, dx \\ & = \textstyle\sum\limits _ {k=1}^n \left[ -\dfrac{(-x)^k}{k} \right] _ 0^1 \\ & = \textstyle\sum\limits _ {k=1}^n \dfrac{(-1)^{k-1}}{k} \end{align}\] また \[\begin{align} T _ n & = \textstyle\sum\limits _ {k=1}^n (-1)^{k-1} \left( \dfrac{1}{k} -\dfrac{1}{k+1} \right) \\ & = \textstyle\sum\limits _ {k=1}^n \dfrac{(-1)^{k-1}}{k} +\textstyle\sum\limits _ {k=2}^{n+1} \dfrac{(-1)^{k-1}}{k} \end{align}\] これらを用いれば \[\begin{align} T _ n -2S _ n & = \textstyle\sum\limits _ {k=2}^{n+1} \dfrac{(-1)^{k-1}}{k} -\textstyle\sum\limits _ {k=1}^n \dfrac{(-1)^{k-1}}{k} \\ & = \underline{\dfrac{(-1)^n}{n+1}-1} \end{align}\]
(3)
(1) の結果について, \(\displaystyle\lim _ {n \rightarrow \infty} \dfrac{1}{n+1} =0\) なので, はさみうちの原理より \[\begin{align} \displaystyle\lim _ {n \rightarrow \infty} & \left| S _ n -\displaystyle\int _ 0^1 \dfrac{1}{1+x} \, dx \right| = 0 \\ \text{∴} \quad \displaystyle\lim _ {n \rightarrow \infty} S _ n & = \displaystyle\int _ 0^1 \dfrac{1}{1+x} \, dx \\ & = \big[ \log (1+x) \big] _ 0^1 = \log 2 \end{align}\]
(2) より \[\begin{align} \displaystyle\lim _ {n \rightarrow \infty} T _ n & = \displaystyle\lim _ {n \rightarrow \infty} \left\{ 2S _ n +\dfrac{(-1)^n}{n+1}-1 \right\} \\ & = 2 \log 2 +0 -1 = \underline{2 \log 2 -1} \end{align}\]