Bài 104560

Question

Chứng minh rằng:

$\frac{1}{1+(n+1)^{2}}<\arctan \frac{1}{n^{2}+n+1}<\frac{1}{1+n^{2}},\forall n\in N$

score 0 · Answer 1

Xét

$f(x)=\arctan x,x \in [n,n+1]$
Theo định lý Lagrange:

$\exists c \in (n,n+1)$

$f(n+1)-f(n)=f'(c)(n+1-n)$

$\Rightarrow \arctan (n+1)-\arctan n=\frac{1}{1+c^{2}}$
Mà:

$\arctan a-\arctan b=\arctan (\frac{a-b}{1+ab})(1)$

$\Rightarrow \arctan \frac{1}{n^{2}+n+1}=\frac{1}{1+c^{2}}$
Hơn nữa:

$n<c<n+1$

$\Leftrightarrow n^{2}<c^{2}<(n+1)^{2}$

$\Leftrightarrow 1+n^{2}<1+c^{2}<1+(n+1)^{2}$

$\Leftrightarrow \frac{1}{1+(n+1)^{2}}<\frac{1}{1+c^{2}}<\frac{1}{1+n^{2}}$
Vậy:

$\frac{1}{1+(n+1)^{2}}<\arctan \frac{1}{n^{2}+n+1}<\frac{1}{1+n^{2}}$

$\Rightarrow$ (ĐPCM)
(Chứng minh (1):

Ta có : $\tan(\arctan a-\arctan b)=\frac{\tan(\arctan a)-\tan(\arctan b)}{1+\tan(\arctan a).\tan(\arctan b)}=\frac{a-b}{1+ab}$

$\Rightarrow\arctan a-\arctan b=\arctan(\frac{a-b}{1+ab})$ )

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