Bac à sable

Existence de la constance
$$n! = \left(\frac{n}{e}\right)^{n} \times \sqrt{2\pi \times n}$$

$$\Longleftrightarrow n! = n^{n} \times \frac{1}{e^{n}} \times \sqrt{2\pi} \times \sqrt{n}$$

$$\Longleftrightarrow n! \times \frac{1}{n^{n}} \times e^{n} \times \frac{1}{\sqrt{n}} = \sqrt{2\pi}$$

$$\Longleftrightarrow \frac{e^{n} \times n!}{n^{n} \times \sqrt{n}} = \sqrt{2\pi}$$

$$f(n)=\frac{e^{n} \times n!}{n^{n} \times \sqrt{n}}$$

$$u_{n} = ln\left[f(n+1)\right] - ln\left[f(n)\right] = ln\left(\frac{e^{n+1} \times (n+1)!}{(n+1)^{n+1} \times \sqrt{n+1}}\right) - ln\left(\frac{e^{n} \times n!}{n^{n} \times \sqrt{n}}\right)$$

$$= ln\left(\frac{\frac{e^{n+1} \times (n+1)!}{(n+1)^{n+1} \times \sqrt{n+1}}}{\frac{e^{n} \times n!}{n^{n} \times \sqrt{n}}}\right) = ln\left(\frac{e^{n+1}\times (n+1)! \times n^{n} \times \sqrt{n}}{(n+1)^{n+1} \times \sqrt{n+1} \times e^{n} \times n!}\right)$$

$$= ln\left(\frac{e^{n+1} \times (n+1)! \times n^{n} \times \sqrt{n}}{e^{n}\times n! \times (n+1)^{n+1} \times \sqrt{n+1}}\right) $$

$$= ln\left(\frac{e^{n+1-n}\times (n+1) \times n! \times n^{n} \times \sqrt{\frac{n}{n+1}}}{n! \times (n+1)^{n+1}}\right)$$

$$= ln\left(\frac{e \times (n+1) \times n^{n} \times n^{\frac{1}{2}}}{(n+1)^{n+1}\times (n+1)^{\frac{1}{2}}}\right)$$

$$= ln\left(\frac{e \times n^{n+\frac{1}{2}}}{(n+1)^{n+\frac{1}{2}}}\right)$$

$$= 1 + ln\left(\frac{n^{n+\frac{1}{2}}}{(n+1)^{n+\frac{1}{2}}}\right)$$

$$= 1 + ln\left[\left(\frac{n}{n+1}\right)^{n+\frac{1}{2}}\right]$$

$$= 1 + ln\left[\left(\frac{n+1}{n}\right)^{-n-\frac{1}{2}}\right]$$

$$= 1 - \left(n+\frac{1}{2}\right)ln\left(\frac{n+1}{n}\right)$$

$$= 1 - \left(n+\frac{1}{2}\right)ln\left(1 + \frac{1}{n}\right)$$

On sait que $$ln(1+x) \sim x - \frac{x^{2}}{2} + \circ x^{3}$$

donc $$ln\left(1 + \frac{1}{n}\right) \sim \frac{1}{n} - \frac{\left(\frac{1}{n}\right)^{2}}{2} + \circ \left(\frac{1}{n}\right)^{3}$$

Donc $$u_{n} \sim 1 - \left(n+\frac{1}{2}\right)\left[\frac{1}{n} - \frac{\left(\frac{1}{n}\right)^{2}}{2} + \circ \left(\frac{1}{n}\right)^{3}\right]$$

$$\sim 1 - \left(n+\frac{1}{2}\right)\times \frac{1}{n} + \left(n+\frac{1}{2}\right)\frac{\left(\frac{1}{n}\right)^{2}}{2} - \left(n+\frac{1}{2}\right)\circ \left(\frac{1}{n}\right)^{3}$$

$$\sim 1 - \frac{n}{n} - \frac{1}{2n} + \frac{n}{2n^{2}} + \frac{1}{4n^{2}}- \left(n+\frac{1}{2}\right)\circ \left(\frac{1}{n}\right)^{3}$$

$$\sim 1 - 1 - \frac{1}{2n} + \frac{1}{2n} + \frac{1}{4n^{2}}- \left(n+\frac{1}{2}\right)\circ \left(\frac{1}{n}\right)^{3}$$

$$\sim \frac{1}{4n^{2}}- \left(n+\frac{1}{2}\right)\circ \left(\frac{1}{n}\right)^{3}$$

$$\sim \frac{1}{4n^{2}} - \circ \frac{1}{n^{2}}$$

Valeur de la constante
$$w_{n} = \int_{0}^{\frac{\pi}{2}}(sin(x))^{n}dx$$

$$w_{0} = \int_{0}^{\frac{\pi}{2}}(sin(x))^{0}dx = \int_{0}^{\frac{\pi}{2}}1dx = [x]_{0}^{\frac{\pi}{2}} = \frac{\pi}{2}$$

$$w_{1} = \int_{0}^{\frac{\pi}{2}}(sin(x))dx = [-cos(x)]_{0}^{\frac{\pi}{2}} = cos(0) - cos\left(\frac{\pi}{2}\right) = 1$$

$$w_{n+1} - w_{n} = \int_{0}^{\frac{\pi}{2}}(sin(x))^{n+1}dx - \int_{0}^{\frac{\pi}{2}}(sin(x))^{n}dx$$

$$= \int_{0}^{\frac{\pi}{2}} (sin(x))^{n+1} - (sin(x))^{n}dx = \int_{0}^{\frac{\pi}{2}} sin(x)^{n}(sin(x)-1)dx$$

Or, sur $$\left[0;\frac{\pi}{2}\right]$$, $$ sin(x)^{n}(sin(x)-1)dx < 0$$ donc $$w_{n+1} - w_{n} < 0$$

Donc la suite $$w_{n}$$ est décroissante.

$$w_{n} = \int_{0}^{\frac{\pi}{2}}(sin(x))^{n}dx = \int_{0}^{\frac{\pi}{2}} sin(x)^{n-2}sin(x)^{2}dx$$

$$= \int_{0}^{\frac{\pi}{2}} sin(x)^{n-2}(1-cos(x)^{2}) dx$$

$$= w_{n-2} - \int_{0}^{\frac{\pi}{2}}sin(x)^{n-2}cos(x)^{2}dx$$

$$[u\times v]_{0}^{\frac{\pi}{2}} = \int_{0}^{\frac{\pi}{2}}u\times v' + \int_{0}^{\frac{\pi}{2}}u'\times v$$

donc $$\int_{0}^{\frac{\pi}{2}}u\times v' = [u\times v]_{0}^{\frac{\pi}{2}} - \int_{0}^{\frac{\pi}{2}}u'\times v$$

On pose $$u = cos(x)$$ donc $$v' = sin(x)^{n-2}cos(x)$$

donc $$u' = -sin(x)$$ et $$v = \frac{sin(x)^{n-1}}{n-1}$$

Donc $$\int_{0}^{\frac{\pi}{2}}sin(x)^{n-2}cos(x)^{2}dx = \left[\frac{cos(x)sin(x)^{n-1}}{n-1}\right]_{0}^{\frac{\pi}{2}} - \int_{0}^{\frac{\pi}{2}}\frac{-sin(x)^{n}}{n-1}dx$$

$$= - \int_{0}^{\frac{\pi}{2}}\frac{-sin(x)^{n}}{n-1}dx = \frac{w_{n}}{n-1}$$

Donc $$w_{n} = w_{n-2} - \frac{w_{n}}{n-1}$$

$$\Longleftrightarrow w_{n} + \frac{w_{n}}{n-1} = w_{n-2} \Longleftrightarrow w_{n}\left(1 + \frac{1}{n-1}\right) = w_{n-2}$$

$$\Longleftrightarrow w_{n} = \frac{w_{n-2}}{1 + \frac{1}{n-1}}  = \frac{w_{n-2}}{\frac{n}{n-1}}$$

$$\Longleftrightarrow w_{n} = \frac{(n-1)}{n}w_{n-2}(R_{1})$$

2k
$$w_{2k} = \frac{2k-1}{2k}w_{2k-2}$$

$$w_{2k-2} = \frac{2k-3}{2k-2}w_{2k-4}$$

donc $$w_{2k} = \frac{2k-1}{2k}\times\frac{2k-3}{2k-2}w_{2k-4}$$

donc $$w_{2k} = \frac{2k-1}{2k}\times\frac{2k-3}{2k-2}\times ... \times \frac{1}{2} \times \frac{\pi}{2}$$

donc $$w_{2k} = \frac{2k}{2k} \times \frac{2k-1}{2k}\times\frac{2k-3}{2k-2}\times ... \times \frac{1}{2} \times \frac{\pi}{2}$$

donc $$w_{2k} = \frac{(2k)!}{2^{2k}\times (k!)^{2}}\times \frac{\pi}{2}$$