Bài 112038

Question

Cho $I_n = \int\limits_{0}^{\frac{\pi}{2} } \sin ^n xdx, n \in N$
a) Chứng minh rằng : $\sqrt{\frac{\pi }{2(n+1)} } < I_n < \sqrt{\frac{\pi}{2n} }$
b) Từ đó suy ra rằng : $ \mathop {\lim }\limits \frac{2.4.6...(2n)}{3.5.7...(2n-1)\sqrt{2n+1} } = \frac{\pi}{2}.$

hoàng anh thọ · Answer 1 · 7/16/2012 8:47:53 AM

a) Đặt $\begin{cases}u= \sin ^{n+1}x \\ dv=\sin xdx \end{cases} \Rightarrow \begin{cases}du=(n+1) \sin ^n x.\cos xdx \\ v= - \cos x \end{cases} $
$\Rightarrow I_{n=2} = \int\limits_{0}^{\frac{\pi}{2} } \sin ^{n+2} xdx = \int\limits_{0}^{\frac{\pi}{2} } \sin ^{n+1}x \sin x.dx$
           $= - \cos x. \sin ^{n+1} x \left| \begin{array}{l}
\frac{\pi }{2}\\
0
\end{array} \right. + (n+1) \int\limits_{0}^{\frac{\pi}{2} } \sin ^n x \cos ^2 xdx$
           $= (n+1) \int\limits_{a0}^{\frac{\pi}{2} } \sin ^nx ( 1- \sin ^2x)dx = (n+1)(I_n - I_{n-2})$
$\Rightarrow ( n+2)I_{n+2} = ( n+1) I_n$
Xét hàm số $ f : N \rightarrow R$
                         $ n \rightarrow f(n) = (n+1)I_n.I_{n+1}$
Ta có: $f(n+1) = (n+2)I_{n+1}.I_{n+2} = (n+1) I_n.I_{n+1} = f(n), \forall n \in N$
$\Rightarrow f(n) = f(1), \forall n \in N$
$\Rightarrow (n+1)I_n.I_{n+1} = \frac{\pi}{2}, \forall n \in N$
Hơn nữa : $I_n > I_{n+1}, \forall n \in N \Rightarrow I^2_{n+1} < I_n.I_{n+1} < I^2_n, \forall n \in N$
                                                $\Rightarrow I^2_{n+1} < \frac{\pi}{2(n+1)} < I^2_n , \forall n \in N$
                                                $\Rightarrow \sqrt{\frac{\pi}{2(n+1)} } < I_n < \sqrt{\frac{\pi}{2n} }, \forall n \in N$

b) Theo bài ... thì $I_n = \begin{cases}\frac{(2k-1)!!}{(2k)!!}.\frac{\pi}{2}, n = 2k   \\ \frac{(2k)!!}{(2k+1)!!}, n = 2k +1 \end{cases}    (k \in N)$
     Mà $I_{2n+1} < I_{2n} < I_{2n-1}, \forall n \in N$
$\Rightarrow \frac{(2n)!!}{(2n+1)!!} < \frac{(2n-1)!!}{(2n)!!}. \frac{\pi}{2} < \frac{(2n-2)!!}{(2n-1)!!}, \forall n \in N$
$\Rightarrow [\frac{(2n)!!}{(2n-1)!!}]^2.\frac{1}{2n+1}<\frac{\pi}{2}< [\frac{(2n)!!}{(2n-1)!!}]^2.\frac{1}{2n}$
$\Rightarrow \sqrt{\frac{\pi}{2} }.\sqrt{\frac{2n}{2n+1} } < \frac{1}{\sqrt{2n+1} }.\frac{(2n)!!}{(2n-1)!!}< \sqrt{\frac{\pi}{2} }$
   Chú ý rằng : $ \mathop {\lim }\limits_{n \to +\infty }\sqrt{\frac{\pi}{2} }.\sqrt{\frac{2}{2n+1} } = \sqrt{\frac{\pi}{2} }$
   Vì vậy : $ \mathop {\lim }\limits_{n \to +\infty }\frac{1}{\sqrt{2n+1} }.\frac{(2n)!!}{(2n-1)!!} = \sqrt{\frac{\pi}{2} }.$

Bài 112038

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Bài 112038

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