Cho $S=\frac{1}{3(\sqrt{1}+\sqrt{2})}+\frac{1}{5(\sqrt{2}+\sqrt{3})}+...+\frac{1}{(2n+1)(\sqrt{n}+\sqrt{n+1})} (n\in \mathbb{N}, n\geqslant 3).$
Chứng minh: $S< \frac{1}{2}.$
Cho $S=\frac{1}{3(\sqrt{1}+\sqrt{2})}+\frac{1}{5(\sqrt{2}+\sqrt{3})}+...+\frac{1}{(2n+1)(\sqrt{n}+\sqrt{n+1})} (n\in \mathbb{N}, n\geqslant 3).$
Chứng minh: $S< \frac{1}{2}.$
Nhận xét:
$$\dfrac{1}{(2n+1)(\sqrt{n}+\sqrt{n+1})}=\dfrac{\sqrt{n+1}-\sqrt{n}}{2n+1}=\dfrac{\sqrt{n+1}-\sqrt{n}}{\sqrt{4n^2+4n+1}}<\dfrac{\sqrt{n+1}-\sqrt{n}}{\sqrt{4n^2+4n}}=\dfrac{\sqrt{n+1}-\sqrt{n}}{2\sqrt{n(n+1)}}=\dfrac{1}{2}\left ( \frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}} \right )$$
Do đó
$$S<\dfrac{1}{2}\left ( \dfrac{1}{\sqrt{1}}-\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{2}}-\dfrac{1}{\sqrt{3}}+\cdots+\dfrac{1}{\sqrt{n}}-\dfrac{1}{\sqrt{n+1}} \right )=\dfrac{1}{2}\left ( 1-\dfrac{1}{\sqrt{n+1}} \right )<\dfrac{1}{2}$$
Bài viết đã được chỉnh sửa nội dung bởi DarkBlood: 27-06-2013 - 14:44
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