Question

cho tổng

S=\(\dfrac{1}{3.\left(1+\sqrt{2}\right)}\)+\(\dfrac{1}{5.\left(\sqrt{2}+\sqrt{3}\right)}\)+......+\(\dfrac{1}{97.\left(\sqrt{48}+\sqrt{49}\right)}\)

So sánh vs \(\dfrac{3}{7}\)

Answer 1

TQ:\(S_n=\dfrac{1}{\left(n+n+1\right)\left(\sqrt{n}+\sqrt{n+1}\right)}=\dfrac{\sqrt{n+1}-\sqrt{n}}{n+\left(n+1\right)}\)

Mà theo AM-GM:\(n+\left(n+1\right)\ge2\sqrt{n\left(n+1\right)}\)

\(\Rightarrow S_n\le\dfrac{\sqrt{n+1}-\sqrt{n}}{2\sqrt{n\left(n+1\right)}}=\dfrac{1}{2}\left(\dfrac{1}{\sqrt{n}}-\dfrac{1}{\sqrt{n+1}}\right)\)

Áp dụng:\(S< \dfrac{1}{2}\left(1-\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{2}}-\dfrac{1}{\sqrt{3}}+...+\dfrac{1}{\sqrt{48}}-\dfrac{1}{\sqrt{49}}\right)\)

\(=\dfrac{1}{2}\left(1-\dfrac{1}{7}\right)=\dfrac{6}{14}=\dfrac{3}{7}\)

Answer 2

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