بين المتساويات التالية :
$\displaystyle{\displaylines{\forall n \in \mathbb{N}^{*}}}$ :
$\displaystyle{\displaylines{\sum_{k=1}^{n} \frac{1}{k (k+1)} = \frac{n}{n+1}}}$
$\displaystyle{\displaylines{\sum_{k=1}^{n} k k! = (n+1)! - 1}}$
$\displaystyle{\displaylines{\sum_{k=1}^{n} \ln\left(1+\frac{1}{k}\right) = \ln(1+n)}}$
$\displaystyle{\displaylines{\sum_{k=1}^{n} \arctan\left(\frac{1}{k^2 + k + 1}\right) = \arctan(n+1) - \frac{\pi}{4}}}$
تذكير :نقول أن المتسلسلة
$\displaystyle{\displaylines{S_n = \sum_{k=1}^{n} w_k}}$ تلسكوبية Télescopique إذا وفقط إذا وجدت متتالية
$\displaystyle{\displaylines{u_n}}$ بحيث :
$\displaystyle{\displaylines{S_n = \sum_{k=1}^{n} (u_{k+1} - u_k) }}$ وفي هذه الحالة لدينا :
$\displaystyle{\displaylines{ S_n = u_{n+1} - u_1 }}$
راجع
رمز المجموع والضرب
لدينا :
$\displaystyle{\displaylines{\begin{array}{rcl}\displaystyle \sum_{k=1}^{n} \frac{1}{k (k+1)} & = & \displaystyle \sum_{k=1}^{n} \left( \frac{1}{k} - \frac{1}{k+1} \right) \\ & = & \displaystyle \sum_{k=1}^{n} \frac{1}{k} - \sum_{k=1}^{n} \frac{1}{k+1} \\ & = & \displaystyle \sum_{k=1}^{n} \frac{1}{k} - \sum_{p=2}^{n+1} \frac{1}{p} \\ & = & \displaystyle \sum_{k=1}^{n} \frac{1}{k} - \sum_{k=2}^{n+1} \frac{1}{k} \\ & = & 1 - \dfrac{1}{n+1} \\ & = & \dfrac{n}{n+1}\end{array}}}$إذن :
$\displaystyle{\displaylines{\forall n \in \mathbb{N}^{*} \, : \, \sum_{k=1}^{n} \frac{1}{k (k+1)} = \frac{n}{n+1}}}$
لدينا :
$\displaystyle{\displaylines{\begin{array}{rcl}\displaystyle \sum_{k=1}^{n} k k! & = & \displaystyle \sum_{k=1}^{n} ((k+1)! - k! ) \\ & = & \displaystyle \sum_{k=1}^{n} (k+1)! - \sum_{k=1}^{n} k! \\ & = & \displaystyle \sum_{p=2}^{n+1} p! - \sum_{k=1}^{n} k! \\ & = & \displaystyle \sum_{k=2}^{n+1} k! - \sum_{k=1}^{n} k! \\ & = & (n+1)!-1\end{array}}}$إذن :
$\displaystyle{\displaylines{\forall n \in \mathbb{N}^{*} \, : \, \sum_{k=1}^{n} k k! = (n+1)! - 1}}$
لدينا :
$\displaystyle{\displaylines{\begin{array}{rcl}\displaystyle \sum_{k=1}^{n} \ln\left( 1+\frac{1}{k} \right) & = & \displaystyle \sum_{k=1}^{n} \ln\left(\frac{k+1}{k} \right) \\ & = & \displaystyle \sum_{k=1}^{n} ( \ln(k+1) - \ln(k) ) \\ & = & \displaystyle \sum_{k=1}^{n} \ln(k+1) - \sum_{k=1}^{n} \ln(k) \\ & = & \displaystyle \sum_{p=2}^{n+1} \ln(p) - \sum_{k=1}^{n} \ln(k) \\ & = & \displaystyle \sum_{k=2}^{n+1} \ln(k) - \sum_{k=1}^{n} \ln(k) \\ & = & \ln(n+1)-\ln(1) \\ & = & \ln(n+1)\end{array}}}$إذن :
$\displaystyle{\displaylines{\forall n \in \mathbb{N}^{*} \, : \, \sum_{k=1}^{n} \ln\left(1+\frac{1}{k}\right) = \ln(1+n)}}$
لنبين أن :
$\displaystyle{\displaylines{\forall k \in \mathbb{N} \ : \ \arctan\left(\frac{1}{k^2 + k + 1}\right) = \arctan(k+1) - \arctan(k)}}$لدينا :
$\displaystyle{\displaylines{\tan(x-y) = \dfrac{\tan(x) - \tan(y)}{1 + \tan(x)\tan(y)} \tag{1}}}$بحيث :
$\displaystyle{\displaylines{x,y,(x-y) \in \mathbb{R} \smallsetminus \left\{ \frac{\pi}{2}+k \pi \ \middle| \ k \in \mathbb{Z} \right\}}}$ليكن
$\displaystyle{\displaylines{u, v \geq 0}}$ بحيث :
$\displaystyle{\displaylines{\left\{ \begin{array}{cl}x & = \ \arctan(u) \\y & = \ \arctan(v)\end{array} \right.}}$لدينا إذن :
$\displaystyle{\displaylines{0 \leq x,y \lt \dfrac{\pi}{2} \tag{2}}}$لدينا :
$\displaystyle{\displaylines{\left\{ \begin{array}{cl}\forall t \in \mathbb{R} & : \ \tan(\arctan(t))=t \\ \\\forall t \in \left] -\dfrac{\pi}{2}, \dfrac{\pi}{2} \right[ & : \ \arctan(\tan(t))=t\end{array} \right.}}$وبالتالي لدينا :
$\displaystyle{\displaylines{\left\{ \begin{array}{cl}\tan(x)=\tan(\arctan(u))=u & \\\tan(y)=\tan(\arctan(v))=v & \end{array} \right.}}$رجوعا للعلاقة
$\displaystyle{\displaylines{(1)}}$ لدينا :
$\displaystyle{\displaylines{\tan(x-y) = \dfrac{u - v}{1 + u v}}}$وبالتالي لدينا :
$\displaystyle{\displaylines{\arctan(\tan(x-y)) = \arctan\left(\dfrac{u - v}{1 + u v}\right)}}$من خلال العلاقة
$\displaystyle{\displaylines{(2)}}$ لدينا :
$\displaystyle{\displaylines{-\dfrac{\pi}{2} < x - y < \dfrac{\pi}{2}}}$وبالتالي فإنه لدينا :
$\displaystyle{\displaylines{\arctan(\tan(x-y)) = x-y = \arctan(u) - \arctan(v)}}$إذن لدينا الخاصية التالية :
$\displaystyle{\displaylines{\forall (u, v) \in \mathbb{R}_{+}^2 \ : \ \arctan(u) - \arctan(v) = \arctan\left(\dfrac{u - v}{1 + u v}\right)}}$من أجل
$\displaystyle{\displaylines{u=k+1}}$ و
$\displaystyle{\displaylines{v=k}}$ لدينا :
$\displaystyle{\displaylines{\forall k \in \mathbb{N} \ : \ \arctan\left(\frac{1}{k^2 + k + 1}\right) = \arctan(k+1) - \arctan(k)}}$وبالتالي :
$\displaystyle{\displaylines{\sum_{k=1}^{n} \arctan\left(\frac{1}{k^2 + k + 1}\right) = \arctan(n+1) - \arctan(1)}}$إذن :
$\displaystyle{\displaylines{\forall n \in \mathbb{N}^{*} \, : \, \sum_{k=1}^{n} \arctan\left(\frac{1}{k^2 + k + 1}\right) = \arctan(n+1) - \frac{\pi}{4}}}$
