部分積分法を用いれば

$$
\begin{eqnarray*}
S_N & = & \int_{0}^{\pi/2}x^2\cos^{2N}{x}\,dx \\
& = & \int_{0}^{\pi/2} x^2(1-\sin^2{x})\cos^{2N-2}{x}\,dx \\
& = & \int_{0}^{\pi/2} x^2\cos^{2N-2}{x}\,dx-\int_{0}^{\pi/2} x^2\sin^2{x}\cos^{2N-2}{x}\,dx \\
& = & S_{N-1}-\int_{0}^{\pi/2} x^2\sin^2{x}\cos^{2N-2}{x}\,dx
\end{eqnarray*}
$$

である．右辺の第２項については，さらに部分積分法を適用して

$$
\begin{eqnarray*}
& & \int_{0}^{\pi/2} x^2\sin^2{x}\cos^{2N-2}{x}\,dx \\
& = & \int_{0}^{\pi/2}x^2\sin{x}\biggl(\frac{-1}{2N-1}\cos^{2N-1}{x}\biggr)’\,dx \\
& = & \biggl[\frac{-x}{2N-1}\sin{x}\cos^{2N-1}{x}\biggr]_{0}^{\pi/2} + \int_{0}^{\pi/2}(x^2\sin{x})’\cdot \frac{1}{2N-1}\cos^{2N-1}{x}\,dx \\
& = & 0 + \int_{0}^{\pi/2}(2x\sin{x} + x^2\cos{x})\cdot \frac{1}{2N-1}\cos^{2N-1}{x}\,dx \\
& = & \frac{2}{2N-1}\int_{0}^{\pi/2}x\sin{x}\cos^{2N-1}{x}\,dx + \frac{1}{2N-1}\int_{0}^{\pi/2} x^2\cos^{2N}{x}\,dx \\
& = & \frac{2}{2N-1}\int_{0}^{\pi/2}x\biggl(\frac{-1}{2N}\cos^{2N}{x}\biggr)’\,dx + \frac{1}{2N-1}S_N \\
& = & \frac{2}{2N-1}\biggl\{\biggl[-\frac{x}{2N}\cos^{2N}{x}\biggr]_{0}^{\pi/2}+\int_{0}^{\pi/2}\frac{1}{2N}\cos^{2N}{x}\,dx\biggr\} + \frac{1}{2N-1}S_N \\
& = & \frac{2}{2N-1}\left(0 + \frac{1}{2N}W_{2N}\right) + \frac{2}{2N-1}S_N \\
& = & \frac{2}{2N(2N-1)}W_{2N} + \frac{1}{2N-1}S_N
\end{eqnarray*}
$$

が成り立つから

$$
\begin{eqnarray*}
S_N & = & \int_{0}^{\pi/2}x^2\cos^{2N}{x}\,dx \\
& = & S_{N-1}-\int_{0}^{\pi/2} x^2\sin^2{x}\cos^{2N-2}{x}\,dx \\
& = & S_{N-1}-\frac{2}{2N(2N-1)}W_{2N}-\frac{1}{2N-1}S_N \\
\frac{2N}{2N-1}S_N & = & S_{N-1}-\frac{2}{2N(2N-1)}W_{2N} \\
S_N & = & \frac{2N-1}{2N}S_{N-1}-\frac{1}{2N^2}W_{2N}
\end{eqnarray*}
$$

を得る．

$\{S_N\}$ の定義より
\[
S_0 = \int_{0}^{\pi/2}x^2\,dx = \biggl[\frac{1}{3}x^3\biggr]_{0}^{\pi/2} = \frac{\pi^3}{24}
\]
である．また，等式 $(\diamondsuit)$ より，$N \geqq 2$ なる整数 $N$ に対して
\[
S_{N-1} = \frac{2N-3}{2N-2}S_{N-2}-\frac{1}{2(N-1)^2}W_{2N-2}
\]
が成り立つ．ゆえに，Wallis 積分に関する漸化式より得られる
\[
W_{2N} = \frac{2N-1}{2N}W_{2N-2}
\]
に注意すれば

$$
\begin{eqnarray*}
S_N & = & \frac{2N-1}{2N}S_{N-1}-\frac{1}{2N^2}W_{2N} \\
& = & \frac{2N-1}{2N}\left\{\frac{2N-3}{2N-2}S_{N-2}-\frac{1}{2(N-1)^2}W_{2N-2}\right\}-\frac{1}{2N-2}W_{2N} \\
& = & \frac{2N-1}{2N}\cdot\frac{2N-3}{2N-2}S_{N-2}-\frac{2N-1}{2N}\cdot\frac{1}{2(N-1)^2}W_{2N-2}-\frac{1}{2N-2}W_{2N} \\
& = & \frac{2N-1}{2N}\cdot\frac{2N-3}{2N-2}S_{N-2}-\frac{1}{2(N-1)^2}W_{2N}-\frac{1}{2N-2}W_{2N} \\
& = & \frac{2N-1}{2N}\cdot\frac{2N-3}{2N-2}S_{N-2}-\frac{1}{2}W_{2N}\left\{\frac{1}{(N-1)^2} + \frac{1}{N^2}\right\}
\end{eqnarray*}
$$

が成り立つ．これを繰り返し用いると，Wallis 積分 で示したWallis積分の組み合わせ記号を用いた表現
\[
W_{2N} = \frac{{}_{2N}{\rm C}{}_{N}}{4^N} \cdot \frac{\pi}{2}
\]
により

$$
\begin{eqnarray*}
S_N & = & \frac{2N-1}{2N}\cdot\frac{2N-3}{2N-2}S_{N-2}-\frac{1}{2}W_{2N}\left\{\frac{1}{(N-1)^2} + \frac{1}{N^2}\right\} \\
& = & \cdots \\
& = & \frac{2N-1}{2N}\cdot\frac{2N-3}{2N-2}\cdots\frac{3}{4}\cdot\frac{1}{2}S_{0}-\frac{1}{2}W_{2N}\left\{\frac{1}{1^2} + \frac{1}{2^2} + \cdots + \frac{1}{(N-1)^2} + \frac{1}{N^2}\right\} \\
& = & \frac{{}_{2N}{\rm C}_{N}}{4^N}S_0-\frac{1}{2}W_{2N}\sum_{n=1}^{N}\frac{1}{n^2} \\
& = & \frac{{}_{2N}{\rm C}_{N}}{4^N}\cdot\frac{\pi^3}{24}-\frac{1}{2} \cdot \frac{{}_{2N}{\rm C}_{N}}{4^N}\cdot \frac{\pi}{2}\sum_{n=1}^{N}\frac{1}{n^2} \\
& = & \frac{{}_{2N}{\rm C}_{N}}{4^N}\cdot \frac{\pi}{4}\left(\frac{\pi^2}{6}-\sum_{n=1}^{N}\frac{1}{n^2}\right)
\end{eqnarray*}
$$

\[
f(x) = \sin{x}-\frac{2}{\pi}x
\]
とおくと
\[
f'(x) = \cos{x}-\frac{2}{\pi}
\]
である．ここで，$f'(x) = g(x)$ とおけば，$g(x)$ は $0 \leqq x \leqq \pi/2$ で連続かつ微分可能であって
\[
g(0) = 1-\frac{2}{\pi} > 0, \quad g\left(\frac{\pi}{2}\right) = -\frac{2}{\pi} < 0
\]
が成り立つから，中間値の定理より，$g(c) = 0$ かつ $0 < c < \pi/2$ を満たす実数 $c$ が存在する．すなわちこの $c$ は $f'(c) = 0$ を満たし，極値をとる候補となり得る．また
\[
f'{}'(c) = -\sin{c} < 0
\]
が成り立つから，$f(x)$ は $x=c$ で極大値をとる．したがって増減表は以下のとおり．

\[
\begin{array}{c||c|c|c|c|c} \hline
x&0&\cdots&c&\cdots&\pi/2 \\ \hline
y’& &+&0&-& \\ \hline
y&0&\nearrow& f(c) &\searrow & 0\\ \hline
\end{array}
\]

ゆえに $0 \leqq x \leqq \pi/2$ で
\[
\sin{x}-\frac{2}{\pi}x = f(x) \geqq 0
\]
であるから
\[
\sin{x} \geqq \frac{2}{\pi}x
\]
を得る．

図形的な意味から考えると判りやすい．

$y = \sin{x}$ のグラフと，２点 $(0,0)$，$(\pi/2,1)$ を通る直線を引けば，不等式はそれらの位置関係に対応する．

補題 4：Jordanの不等式 $(\heartsuit)$ より，$0 \leqq x \leqq \pi/2$ において
\[
0 \leqq x \leqq \frac{\pi}{2}\sin{x}
\]
が成り立つから
\[
0 \leqq x^2\cos^{2N}{x} \leqq \frac{\pi^2}{4}\sin^2{x}\cos^{2N}{x}
\]
である．したがって
$$
\begin{eqnarray*}
0 ~ \leqq ~ S_N & = & \int_{0}^{\pi/2}x^2\cos^{2N}{x}\,dx \\
& \leqq & \int_{0}^{\pi/2}\frac{\pi^2}{4}\sin^{2}{x}\cos^{2N}{x}\,dx \\
& = & \frac{\pi^2}{4}\int_{0}^{\pi/2}(\cos^{2N}{x}-\cos^{2N+2}{x})\,dx \\
& = & \frac{\pi^2}{4} (W_{2N}-W_{2N+2}) \\
& = & \frac{\pi^2}{4}\left(W_{2N}-\frac{2N+1}{2N+2}W_{2N} \right)\\
& = & \frac{\pi^2}{4}\cdot \frac{1}{2N+2}\cdot \frac{{}_{2N}{\rm C}_{N}}{4^N}\cdot \frac{\pi}{2} \\
& = & \frac{1}{2N+2}\cdot \frac{{}_{2N}{\rm C}_{N}}{4^N}\cdot \frac{\pi^3}{8}
\end{eqnarray*}
$$

を得る．

不等式 $(\spadesuit)$ に 等式 $(\clubsuit)$ の結果を代入して変形すると
\[
0 \leqq S_N \leqq \frac{1}{2N+2}\cdot \frac{{}_{2N}{\rm C}_{N}}{4^N}\cdot \frac{\pi^3}{8}
\]
\[
0 \leqq \frac{{}_{2N}{\rm C}_{N}}{4^N}\cdot \frac{\pi}{4}\left(\frac{\pi^2}{6}-\sum_{n=1}^{N}\frac{1}{n^2}\right) \leqq \frac{1}{2N+2}\cdot \frac{{}_{2N}{\rm C}_{N}}{4^N}\cdot \frac{\pi^3}{8}
\]
\[
0 \leqq \frac{\pi^2}{6}-\sum_{n=1}^{N}\frac{1}{n^2} \leqq \frac{1}{2N+2} \cdot \frac{\pi^2}{2}
\]
\[
– \frac{\pi^2}{6} \leqq -\sum_{n=1}^{N}\frac{1}{n^2} \leqq \frac{1}{2N+2} \cdot \frac{\pi^2}{2}-\frac{\pi^2}{6}
\]
\[
\frac{\pi^2}{6} + \frac{1}{2N+2} \cdot \frac{\pi^2}{2} \leqq \sum_{n=1}^{N}\frac{1}{n^2} \leqq \frac{\pi^2}{6}
\]

を得る．ここで
\[
\lim_{N \to \infty}\left(\frac{\pi^2}{6} + \frac{1}{2N+2}\right) = \frac{\pi^2}{6}
\]
であるから，はさみうちの原理より
\[
\sum_{n=1}^{\infty} \frac{1}{n^2} = \frac{\pi^2}{6}
\]
を得る．

\[
\sum_{n=1}^{\infty}\frac{1}{n^2} = \frac{1}{1^2} + \frac{1}{2^2} + \frac{1}{3^2} + \cdots
\]
を奇数と偶数に分けると
\[
\sum_{n=1}^{\infty}\frac{1}{n^2} = \sum_{n=1}^{\infty}\frac{1}{(2n-1)^2} + \sum_{n=1}^{\infty}\frac{1}{(2n)^2} = \sum_{n=1}^{\infty}\frac{1}{(2n-1)^2} + \frac{1}{4}\sum_{n=1}^{\infty}\frac{1}{n^2}
\]
より
\[
\sum_{n=1}^{\infty}\frac{1}{(2n-1)^2} = \frac{3}{4}\sum_{n=1}^{\infty}\frac{1}{n^2} = \frac{3}{4}\cdot\frac{\pi^2}{6} = \frac{\pi^2}{8}
\]
である．また
\[
\sum_{n=1}^{\infty}\frac{1}{(2n)^2} = \frac{1}{4}\sum_{n=1}^{\infty}\frac{1}{n^2} = \frac{1}{4}\cdot\frac{\pi^2}{6} = \frac{\pi^2}{24}
\]
であるから，題意は示された．

\[
\sum_{n=1}^{\infty}\frac{1}{n^2} = \frac{1}{1^2} + \frac{1}{2^2} + \frac{1}{3^2} + \cdots
\]

を，$4$で割った余りによって４つの級数に分ける：

$$
\begin{eqnarray*}
\sum_{n=1}^{\infty}\frac{1}{n^2}
&=& \sum_{n=1}^{\infty}\frac{1}{(4n-1)^2} + \sum_{n=1}^{\infty}\frac{1}{(4n-2)^2} + \sum_{n=1}^{\infty}\frac{1}{(4n-3)^2} + \sum_{n=1}^{\infty}\frac{1}{(4n)^2} \\
&=& \sum_{n=1}^{\infty}\frac{1}{(4n-1)^2} + \frac{1}{4}\sum_{n=1}^{\infty}\frac{1}{(2n-1)^2} + \sum_{n=1}^{\infty}\frac{1}{(4n-3)^2} + \frac{1}{16}\sum_{n=1}^{\infty}\frac{1}{n^2}
\end{eqnarray*}
$$

ここで，Corollary 1の計算中に現れた結果
\[
\sum_{n=1}^{\infty}\frac{1}{(2n-1)^2} = \frac{\pi^2}{8}
\]
に注意すれば
\[
\sum_{n=1}^{\infty}\frac{1}{(4n-1)^2} + \sum_{n=1}^{\infty}\frac{1}{(4n-3)^2} = \left(\frac{1}{6}-\frac{1}{16 \cdot 6}-\frac{1}{32}\right)\pi^2 = \frac{\pi^2}{8}
\]
である．左辺を整理すると
\[
\sum_{n=1}^{\infty}\frac{1}{(4n-1)^2} + \sum_{n=1}^{\infty}\frac{1}{(4n-3)^2} = 2\sum_{n=1}^{\infty}\frac{16n^2-16n + 5}{(4n-1)^2(4n-3)^2}
\]
であるから
\[
\sum_{n=1}^{\infty}\frac{16n^2-16n + 5}{(4n-1)^2(4n-3)^2} = \frac{1}{2} \cdot \frac{\pi^2}{8} = \frac{\pi^2}{16}
\]
を得る．