Ta có:
\(\dfrac{1}{\left(k+1\right)\sqrt{k}}=\dfrac{\sqrt{k}}{k\left(k+1\right)}=\sqrt{k}\left(\dfrac{1}{k}-\dfrac{1}{k+1}\right)\)
\(=\sqrt{k}\left(\dfrac{1}{\sqrt{k}}+\dfrac{1}{\sqrt{k+1}}\right)\left(\dfrac{1}{\sqrt{k}}-\dfrac{1}{\sqrt{k+1}}\right)\)
\(=\left(1+\dfrac{\sqrt{k}}{\sqrt{k+1}}\right)\left(\dfrac{1}{\sqrt{k}}-\dfrac{1}{\sqrt{k+1}}\right)< 2\left(\dfrac{1}{\sqrt{k}}-\dfrac{1}{\sqrt{k+1}}\right)\)
\(A=\dfrac{1}{2\sqrt{1}}+\dfrac{1}{3\sqrt{2}}+...+\dfrac{1}{\left(n+1\right)\sqrt{n}}\)
\(\Rightarrow A< 2\left[\left(1-\dfrac{1}{\sqrt{2}}\right)+\left(\dfrac{1}{\sqrt{2}}-\dfrac{1}{\sqrt{3}}\right)+...+\left(\dfrac{1}{\sqrt{n}}-\dfrac{1}{\sqrt{n+1}}\right)\right]\)
\(\Rightarrow A< 2\left(1-\dfrac{1}{\sqrt{n+1}}\right)< 2\)
Vậy \(A< 2\) (dpcm)
T đang bận tý , lát giải cho nhé.
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