أحسب نهاية المتسلسلة التالية

نعتبر المتسلسلة التالية $\displaystyle{\displaylines{\forall n \in \mathbb{N}^{*} \quad S_n = \frac{1}{\sqrt{n}} \, \sum_{k=1}^{n} \frac{1}{\sqrt{k}}}}$

1) باستغلالك لتباعد المتسلسلة $\displaystyle{\displaylines{ \sum_{k=1}^{n} \frac{1}{k} }}$ بين أن $\displaystyle{\displaylines{ \sqrt{n} \, S_n}}$ متباعدة.

2) بين أن $\displaystyle{\displaylines{\forall k \in \mathbb{N}^{*} \ : \ \frac{1}{\sqrt{k+1}} \leq 2 \, (\sqrt{k+1} - \sqrt{k}) \leq \frac{1}{\sqrt{k}}}}$

3) استنتج أن $\displaystyle{\displaylines{\forall n \in \mathbb{N}^{*} \ : \ 2 \, \sqrt{1+\frac{1}{n}} - \frac{2}{\sqrt{n}} \leq S_n \leq 2 - \frac{1}{\sqrt{n}}}}$

4) استنتج أن $\displaystyle{\displaylines{S_n}}$ متقاربة وحدد نهايتها

لدينا المتسلسلة $\displaystyle{\displaylines{ \sum_{k=1}^{n} \frac{1}{k} }}$ متباعدة. (راجع حدد تقارب المتتالية التالية)

ولدينا :

$\displaystyle{\displaylines{\forall k \in \mathbb{N}^{*} \, : \, \frac{1}{k} \leq \frac{1}{\sqrt{k}}}}$

$\displaystyle{\displaylines{\forall n \in \mathbb{N}^{*} \, : \, \sum_{k=1}^{n} \frac{1}{k} \leq \sum_{k=1}^{n} \frac{1}{\sqrt{k}}}}$

لدينا : $\displaystyle{\displaylines{\lim_{n\rightarrow + \infty} \sum_{k=1}^{n} \frac{1}{k} = + \infty}}$

إذن :

$\displaystyle{\displaylines{\lim_{n\rightarrow + \infty} \sum_{k=1}^{n} \frac{1}{\sqrt{k}} = \lim_{n\rightarrow + \infty} \sqrt{n} \, S_n = + \infty}}$

ليكن $\displaystyle{\displaylines{ k \in \mathbb{N}^{*} }}$

لدينا :

$\displaystyle{\displaylines{\sqrt{k+1} - \sqrt{k} = \frac{(\sqrt{k+1} - \sqrt{k}) \, (\sqrt{k+1} + \sqrt{k})}{\sqrt{k+1} + \sqrt{k}} = \frac{1}{\sqrt{k+1} + \sqrt{k}} \tag{1}}}$

من جهة أخرى : $\displaystyle{\displaylines{ \sqrt{k} \leq \sqrt{k+1} }}$ إذن :

$\displaystyle{\displaylines{2 \, \sqrt{k} \leq \sqrt{k+1} + \sqrt{k} \leq 2 \, \sqrt{k+1}}}$

$\displaystyle{\displaylines{\frac{1}{2 \, \sqrt{k+1}} \leq \frac{1}{\sqrt{k+1} + \sqrt{k}} \leq \frac{1}{2 \, \sqrt{k}} \tag{2}}}$

من $\displaystyle{\displaylines{(1)}}$ و $\displaystyle{\displaylines{(2)}}$ لدينا :

$\displaystyle{\displaylines{ \forall k \in \mathbb{N}^{*} \, : \, \frac{1}{\sqrt{k+1}} \leq 2 \, (\sqrt{k+1} - \sqrt{k}) \leq \frac{1}{\sqrt{k}}}}$

لدينا :

$\displaystyle{\displaylines{\forall k \in \mathbb{N}^{*} \, : \, 2 \, (\sqrt{k+1} - \sqrt{k}) \leq \frac{1}{\sqrt{k}}}}$

$\displaystyle{\displaylines{\forall n \in \mathbb{N}^{*} \, : \, 2 \, \sum_{k=1}^{n} \sqrt{k+1} - \sqrt{k} \leq \sum_{k=1}^{n} \frac{1}{\sqrt{k}}}}$

$\displaystyle{\displaylines{\forall n \in \mathbb{N}^{*} \, : \, 2 \, (\sqrt{n+1} - 1) \leq \sqrt{n} \, S_n}}$

$\displaystyle{\displaylines{\forall n \in \mathbb{N}^{*} \, : \, 2 \, \sqrt{1+\frac{1}{n}} - \frac{2}{\sqrt{n}} \leq S_n \tag{3}}}$

ولدينا

$\displaystyle{\displaylines{ \forall k \in \mathbb{N}^{*} \, : \, \frac{1}{\sqrt{k+1}} \leq 2 \, (\sqrt{k+1} - \sqrt{k})}}$

$\displaystyle{\displaylines{ \forall n \in \mathbb{N}^{*} \, : \, \sum_{k=1}^{n} \frac{1}{\sqrt{k+1}} \leq 2 \, \sum_{k=1}^{n} (\sqrt{k+1} - \sqrt{k})}}$

$\displaystyle{\displaylines{ \forall n \in \mathbb{N}^{*} \, : \, \sum_{k=2}^{n+1} \frac{1}{\sqrt{k}} \leq 2 \, \sqrt{n+1} - 2}}$

$\displaystyle{\displaylines{ \forall n \in \mathbb{N}^{*} \, : \, \sum_{k=1}^{n+1} \frac{1}{\sqrt{k}} \leq 2 \, \sqrt{n+1} - 1}}$

$\displaystyle{\displaylines{ \forall n \in \mathbb{N}^{*} \, : \, \sum_{k=1}^{n} \frac{1}{\sqrt{k}} = \sqrt{n} \, S_n \leq 2 \, \sqrt{n} - 1}}$

$\displaystyle{\displaylines{ \forall n \in \mathbb{N}^{*} \, : \, S_n \leq 2 - \frac{1}{\sqrt{n}} \tag{4}}}$

من $\displaystyle{\displaylines{(3)}}$ و $\displaystyle{\displaylines{(4)}}$ لدينا :

$\displaystyle{\displaylines{\forall n \in \mathbb{N}^{*} \, : \, 2 \, \sqrt{1+\frac{1}{n}} - \frac{2}{\sqrt{n}} \leq S_n \leq 2 - \frac{1}{\sqrt{n}}}}$

لدينا $\displaystyle{\displaylines{(S_n)_{n \in \mathbb{N}^{*}} }}$ تزايدية ( $\displaystyle{\displaylines{S_{n+1} - S_n = \frac{1}{\sqrt{n+1}} > 0}}$ ) .

ولدينا أيضا : $\displaystyle{\displaylines{\forall n \in \mathbb{N}^{*} \, : \, S_n \leq 2 - \frac{1}{\sqrt{n}} < 2}}$ .

$\displaystyle{\displaylines{(S_n)_{n \in \mathbb{N}^{*}}}}$ تزايدية ومكبورة بـ$\displaystyle{\displaylines{2}}$ إذن متقاربة, وبما أن :

$\displaystyle{\displaylines{\forall n \in \mathbb{N}^{*} \, : \, 2 \, \sqrt{1+\frac{1}{n}} - \frac{2}{\sqrt{n}} \leq S_n \leq 2 - \frac{1}{\sqrt{n}}}}$

وأيضا :

$\displaystyle{\displaylines{\lim_{n \rightarrow + \infty} 2 \, \sqrt{1+\frac{1}{n}} - \frac{2}{\sqrt{n}} = \lim_{n \rightarrow + \infty} 2 - \frac{1}{\sqrt{n}} = 2}}$

فإن :

$\displaystyle{\displaylines{\lim_{n \rightarrow + \infty} S_n = 2}}$