chung minh rang voi moi n lon hon 1 ta co \(\sqrt{2\sqrt{3\sqrt{4...\sqrt{\left(n-1\right)\sqrt{n}}}}}< 3\)
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\(\dfrac{1}{\left(n+1\right)\sqrt{n}+n\sqrt{n+1}}=\dfrac{\left(n+1\right)\sqrt{n}-n\sqrt{n+1}}{\left(n+1\right)^2n-n^2\left(n+1\right)}\)
\(=\dfrac{\left(n+1\right)\sqrt{n}-n\sqrt{n+1}}{n\left(n+1\right)}=\dfrac{1}{\sqrt{n}}-\dfrac{1}{\sqrt{n+1}}\)
Do đó:
\(VT=\dfrac{1}{\sqrt{1}}-\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{2}}-\dfrac{1}{\sqrt{3}}+...+\dfrac{1}{\sqrt{n}}-\dfrac{1}{\sqrt{n+1}}\)
\(VT=1-\dfrac{1}{\sqrt{n+1}}< 1\) (đpcm)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\dfrac{1}{\sqrt{n}.\sqrt{n+1}.\left(\sqrt{n+1}+\sqrt{n}\right)}=\dfrac{1}{\sqrt{n}.\sqrt{n+1}}.\dfrac{1}{\sqrt{n+1}+\sqrt{n}}=\dfrac{1}{\sqrt{n}.\sqrt{n+1}}.\dfrac{\sqrt{n+1}-\sqrt{n}}{n+1-n}=\dfrac{1}{\sqrt{n}.\sqrt{n+1}}.\left(\sqrt{n+1}-\sqrt{n}\right)=\dfrac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n}.\sqrt{n+1}}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Lời giải:
Xét số hạng tổng quát:
\(\frac{1}{(n+1)\sqrt{n}}=\frac{(\sqrt{n+1}-\sqrt{n})(\sqrt{n+1}+\sqrt{n})}{(n+1)\sqrt{n}}<\frac{(\sqrt{n+1}-\sqrt{n}).2\sqrt{n+1}}{(n+1)\sqrt{n}}\)
Hay \(\frac{1}{(n+1)\sqrt{n}}< \frac{2\sqrt{n+1}-\sqrt{n}}{\sqrt{n(n+1)}}=\frac{2}{\sqrt{n}}-\frac{2}{\sqrt{n+1}}\)
Áp dụng vào bài toán:
\(\frac{1}{2\sqrt{1}}+\frac{1}{3\sqrt{2}}+\frac{1}{4\sqrt{3}}+...+\frac{1}{(n+1)\sqrt{n}}< \frac{2}{\sqrt{1}}-\frac{2}{\sqrt{2}}+\frac{2}{\sqrt{2}}-\frac{2}{\sqrt{3}}+\frac{2}{\sqrt{3}}-\frac{2}{\sqrt{4}}+....+\frac{2}{\sqrt{n}}-\frac{2}{\sqrt{n+1}}=2-\frac{2}{\sqrt{n+1}}< 2\)
Ta có đpcm.
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