الهدف من هذا التمرين هو إثبات المتساوية
$\displaystyle{\displaylines{\sum_{n=1}^{+ \infty}\frac{1}{n^2} = \frac{\pi^2}{6}}}$.
ليكن
$\displaystyle{\displaylines{n \in \mathbb{N}}}$ نضع :
$\displaystyle{\displaylines{I_n=\int_{0}^{\frac{\pi}{2}}\cos^{2n}(x)dx}}$ و
$\displaystyle{\displaylines{J_n=\int_{0}^{\frac{\pi}{2}}x^2\cos^{2n}(x)dx}}$1) أحسب $\displaystyle{\displaylines{I_0}}$ و $\displaystyle{\displaylines{J_0}}$.
2) بين ان $\displaystyle{\displaylines{\forall n \in \mathbb{N}^{*} \quad 2n I_n = (2n - 1) I_{n-1}}}$
3) بين ان $\displaystyle{\displaylines{\forall n \in \mathbb{N}^{*} \quad I_n=n(2n-1)J_{n-1}-2n^2J_n}}$
4) استنتج أن $\displaystyle{\displaylines{\forall n \in \mathbb{N}^{*} \quad \frac{1}{n^2}=2\left(\frac{J_{n-1}}{I_{n-1}}-\frac{J_n}{I_n}\right)}}$
5) إستنتج أن $\displaystyle{\displaylines{\forall n \in \mathbb{N}^{*} \quad \sum_{k=1}^{n}\frac{1}{k^2} = \frac{\pi^2}{6}- 2\cdot\frac{J_n}{I_n}}}$
6) بين أن $\displaystyle{\displaylines{\forall x \in [0, \frac{\pi}{2}] \quad x \leq \frac{\pi}{2} \sin(x)}}$
7) استنتج ان : $\displaystyle{\displaylines{\forall n \in \mathbb{N} \quad 0 \leq J_n \leq \frac{\pi^2}{4} \, \frac{1}{2n+2} \, I_n}}$
8) استنتج ان $\displaystyle{\displaylines{\sum_{n=1}^{+ \infty}\frac{1}{n^2} = \frac{\pi^2}{6}}}$
$\displaystyle{\displaylines{\begin{array}{rcl}I_0 & = & \displaystyle\int_{0}^{\frac{\pi}{2}} dx \\ \\& = & \dfrac{\pi}{2} \end{array}}}$
$\displaystyle{\displaylines{\begin{array}{rcl}J_0 & = & \displaystyle\int_{0}^{\frac{\pi}{2}} x^2 dx \\ \\& = & \dfrac{\pi^3}{24} \end{array}}}$
لدينا
$\displaystyle{\displaylines{I_{n} = \int_{0}^{\frac{\pi}{2}} \cos^{2n}(x)dx}}$نقوم بالمكاملة بأجزاء :
$\displaystyle{\displaylines{\begin{array}{rcl}u(x)=\cos^{2n-1}(x) & \longrightarrow & u^{'}(x) = -(2n-1)\sin(x)\cos^{2(n-1)}(x) \\v^{'}(x) = \cos(x) & \longrightarrow & v(x)=\sin(x)\end{array}}}$
وبالتالي لدينا :
$\displaystyle{\displaylines{\begin{array}{rcl}I_{n} & = & \left[ \sin(x) \cos^{2n-1}(x) \right]_{0}^{\pi/2} + (2n-1) \displaystyle\int_{0}^{\frac{\pi}{2}} \sin^{2}(x)\cos^{2(n-1)}(x) dx \\ \\& = & 0 + (2n-1) \displaystyle\int_{0}^{\frac{\pi}{2}} (1-\cos^{2}(x)) \cos^{2(n-1)}(x) dx \\ \\& = & (2n-1) (I_{n-1} - I_{n}) \end{array}}}$
وبالتالي فإنه لدينا :
$\displaystyle{\displaylines{\forall n \in \mathbb{N}^{*} \quad 2 n I_{n} = (2 n - 1) I_{n-1}}}$
لدينا
$\displaystyle{\displaylines{I_{n} = \int_{0}^{\frac{\pi}{2}} \cos^{2n}(x)dx \, , \quad J_{n} = \int_{0}^{\frac{\pi}{2}} x^2 \cos^{2n}(x)dx}}$نقوم بالمكاملة بأجزاء :
$\displaystyle{\displaylines{\begin{array}{rcl}u(x)=\cos^{2n}(x) & \longrightarrow & u^{'}(x) = -2n \, \sin(x)\cos^{2n-1}(x) \\v^{'}(x) = 1 & \longrightarrow & v(x)=x\end{array}}}$
$\displaystyle{\displaylines{\begin{array}{rcl}I_{n} & = & \left[ x \cos^{2n}(x) \right]_{0}^{\pi/2} + 2n \displaystyle\int_{0}^{\frac{\pi}{2}} x \, \sin(x)\cos^{2n-1}(x) dx \\ \\& = & 2n \displaystyle\int_{0}^{\frac{\pi}{2}} x \, \sin(x)\cos^{2n-1}(x) dx \end{array}}}$
نقوم بالمكاملة بأجزاء مرة أخرى :
$\displaystyle{\displaylines{\begin{array}{rcl}u(x)=\sin(x) \cos^{2n-1}(x) & \longrightarrow & u^{'}(x) = \cos^{2n}(x) - (2n-1) \, \sin^{2}(x) \cos^{2(n-1)}(x) \\v^{'}(x) = x & \longrightarrow & v(x)=\frac{x^2}{2}\end{array}}}$
وبالتالي لدينا :
$\displaystyle{\displaylines{\begin{array}{rcl}I_{n} & = & 2 n \, \left[ \frac{x^2}{2} \sin(x) \cos^{2n-1}(x) \right]_{0}^{\pi/2} - n \displaystyle\int_{0}^{\frac{\pi}{2}} x^2 \cos^{2n}(x) dx \\ \\& \, & + (2n-1) n \displaystyle\int_{0}^{\frac{\pi}{2}} x^2 (1-\cos^{2}(x)) \cos^{2(n-1)}(x) dx \\& = & - n J_{n} + n (2n-1) J_{n-1} - (2n-1)n J_{n} \end{array}}}$
وبالتالي فإنه لدينا
$\displaystyle{\displaylines{\forall n \in \mathbb{N}^{*} \quad I_{n} = n (2n-1) J_{n-1} - 2 n^{2} J_{n}}}$
يجب الإشارة هنا أن
$\displaystyle{\displaylines{\forall n \in \mathbb{N} \,\, I_{n} \neq 0}}$ لأن
$\displaystyle{\displaylines{I_{n} = \int_{0}^{\frac{\pi}{2}} \cos^{2n}(x) dx}}$ والدالة
$\displaystyle{\displaylines{\cos}}$ متصلة وموجبة على المجال
$\displaystyle{\displaylines{[0, \frac{\pi}{2}]}}$ و
$\displaystyle{\displaylines{\cos \neq t \rightarrow 0}}$ على هذا المجال.
ليكن
$\displaystyle{\displaylines{n\in\mathbb{N}^{*}}}$ لدينا :
$\displaystyle{\displaylines{I_{n} = n (2n-1) J_{n-1} - 2n^{2} J_{n}}}$نقوم بالقسمة على
$\displaystyle{\displaylines{I_{n-1}}}$ :
$\displaystyle{\displaylines{\frac{I_{n}}{I_{n-1}} = n (2n-1) \frac{J_{n-1}}{I_{n-1}} - 2n^{2} \frac{J_{n}}{I_{n-1}} \quad (\star)}}$لدينا :
$\displaystyle{\displaylines{2n I_n = (2n - 1) I_{n-1}}}$وبالتالي
$\displaystyle{\displaylines{\frac{I_{n}}{I_{n-1}} = \frac{2n-1}{2n}}}$ و
$\displaystyle{\displaylines{I_{n-1} = \frac{2n}{2n-1} I_{n}}}$نقوم بالتعويض في
$\displaystyle{\displaylines{(\star)}}$ :
$\displaystyle{\displaylines{\frac{2n-1}{2n} = n (2n-1) \frac{J_{n-1}}{I_{n-1}} - 2 n^{2} \dfrac{J_{n}}{\dfrac{2n}{2n-1} I_{n}}}}$وبالتالي :
$\displaystyle{\displaylines{\frac{2n-1}{2n} = n (2n-1) \frac{J_{n-1}}{I_{n-1}} - n (2n -1) \dfrac{J_{n}}{I_{n}}}}$إذن
$\displaystyle{\displaylines{\forall n \in \mathbb{N}^{*} \quad \frac{1}{n^2}=2\left(\frac{J_{n-1}}{I_{n-1}}-\frac{J_n}{I_n}\right)}}$
لدينا
$\displaystyle{\displaylines{\forall k \in \mathbb{N}^{*} \quad \frac{1}{k^{2}} = 2 \left( \frac{J_{k-1}}{I_{k-1}} - \frac{J_{k}}{I_{k}} \right)}}$$\displaystyle{\displaylines{\sum_{k=1}^{n} \frac{1}{k^{2}} = 2 \sum_{k=1}^{n} \left( \frac{J_{k-1}}{I_{k-1}} - \frac{J_{k}}{I_{k}} \right)}}$المجموع على اليمين مجموع تليسكوبي :
$\displaystyle{\displaylines{\sum_{k=1}^{n} \frac{1}{k^{2}} = 2 \left( \frac{J_{0}}{I_{0}} - \frac{J_{n}}{I_{n}} \right)}}$رجوعا إلى السؤال الأول لدينا
$\displaystyle{\displaylines{\frac{J_{0}}{I_{0}} = \frac{\pi^2}{12}}}$إذن
$\displaystyle{\displaylines{\sum_{k=1}^{n} \frac{1}{k^{2}} = \frac{\pi^2}{6} - 2 \frac{J_{n}}{I_{n}}}}$
ليكن
$\displaystyle{\displaylines{x \in [0, \frac{\pi}{2}]}}$لدينا دالة
$\displaystyle{\displaylines{\sin}}$ مقعرة (concave) على هذا المجال :
$\displaystyle{\displaylines{\forall x \in [0, \frac{\pi}{2}] \quad \sin^{''}(x) = - \sin(x) \leq 0}}$وبالتالي فإن منحنى الدالة
$\displaystyle{\displaylines{\sin}}$ سيكون فوق منحنى المستقيم المار من النقطتين
$\displaystyle{\displaylines{(0, \sin(0)=0)}}$ و
$\displaystyle{\displaylines{\left(\frac{\pi}{2}, \sin(\frac{\pi}{2}) = 1 \right)}}$معادلة المستقيم المار من هتين النقطتين هو
$\displaystyle{\displaylines{y = \frac{2}{\pi} x}}$وبالتالي لدينا :
$\displaystyle{\displaylines{\forall x \in [0, \frac{\pi}{2}] \quad \frac{2}{\pi} x \leq \sin(x)}}$$\displaystyle{\displaylines{\forall x \in [0, \frac{\pi}{2}] \quad x \leq \frac{\pi}{2} \sin(x)}}$
لدينا :
$\displaystyle{\displaylines{\forall x \in [0, \frac{\pi}{2}] \quad 0 \leq x \leq \frac{\pi}{2} \sin(x)}}$إذن :
$\displaystyle{\displaylines{\forall x \in [0, \frac{\pi}{2}] \quad 0 \leq x^2 \leq \frac{\pi^{2}}{4} \sin^{2}(x)}}$وبالتالي :
$\displaystyle{\displaylines{\forall x \in [0, \frac{\pi}{2}] \quad 0 \leq x^2 \leq \frac{\pi^{2}}{4} (1-\cos^{2}(x))}}$إذن :
$\displaystyle{\displaylines{\forall x \in [0, \frac{\pi}{2}] \quad 0 \leq x^2 \cos^{2n}(x) \leq \frac{\pi^{2}}{4} (1-\cos^{2}(x)) \cos^{2n}(x)}}$وبالتالي :
$\displaystyle{\displaylines{0 \leq \int_{0}^{\frac{\pi}{2}} x^2 \cos^{2n}(x) dx \leq \frac{\pi^{2}}{4} \int_{0}^{\frac{\pi}{2}} (1-\cos^{2}(x)) \cos^{2n}(x) dx}}$إذن :
$\displaystyle{\displaylines{0 \leq J_{n} \leq \frac{\pi^{2}}{4} (I_{n} - I_{n+1})}}$لدينا حسب السؤال الثاني :
$\displaystyle{\displaylines{I_{n+1} = \frac{2n+1}{2(n+1)} I_{n}}}$وبالتالي :
$\displaystyle{\displaylines{\forall n \in \mathbb{N} \quad 0 \leq J_{n} \leq \frac{\pi^{2}}{4} \frac{1}{2n+2} I_{n}}}$
لدينا :
$\displaystyle{\displaylines{\forall n \in \mathbb{N}^{*} \quad \sum_{n=1}^{n}\frac{1}{n^2} = \frac{\pi^2}{6}- 2\cdot\frac{J_n}{I_n}}}$ولدينا أيضا :
$\displaystyle{\displaylines{\forall n \in \mathbb{N} \quad 0 \leq \frac{J_n}{I_n} \leq \frac{\pi^2}{4}\cdot\frac{1}{2n+2}}}$وبما أن
$\displaystyle{\displaylines{\lim_{n \rightarrow +\infty} \frac{\pi^2}{4}\cdot\frac{1}{2n+2}= \lim_{n \rightarrow +\infty} 0 = 0}}$فإن
$\displaystyle{\displaylines{\lim_{n \rightarrow +\infty} \frac{J_n}{I_n} = 0}}$إذن
$\displaystyle{\displaylines{\sum_{n=1}^{+\infty}\frac{1}{n^2} = \frac{\pi^2}{6}}}$ 
