CMR:\(1+\dfrac{1}{2\sqrt{1}}+\dfrac{1}{3\sqrt{2}}+...+\dfrac{1}{\left(n+1\right)\sqrt{n}}< 2\)
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
1 tháng 8 2018
Ta có : \(\sqrt{n+1}-\sqrt{n}=\dfrac{\left(\sqrt{n+1}-\sqrt{n}\right)\left(\sqrt{n+1}+\sqrt{n}\right)}{\sqrt{n+1}+\sqrt{n}}=\dfrac{1}{\sqrt{n+1}+\sqrt{n}}< \dfrac{1}{\sqrt{n}+\sqrt{n}}=\dfrac{1}{2\sqrt{n}}\) ⇒ \(2\left(\sqrt{n+1}-\sqrt{n}\right)< \dfrac{1}{\sqrt{n}}\left(1\right)\)
\(\sqrt{n}-\sqrt{n-1}=\dfrac{\left(\sqrt{n}-\sqrt{n-1}\right)\left(\sqrt{n}+\sqrt{n+1}\right)}{\sqrt{n}+\sqrt{n-1}}=\dfrac{1}{\sqrt{n}+\sqrt{n-1}}>\dfrac{1}{\sqrt{n}+\sqrt{n}}=\dfrac{1}{2\sqrt{n}}\) ⇒ \(2\left(\sqrt{n+1}-\sqrt{n}\right)>\dfrac{1}{\sqrt{n}}\left(2\right)\)
Từ \(\left(1;2\right)\text{⇒ }đpcm\)
nè
10 tháng 9 2017
\(\dfrac{1}{\sqrt{n}.\sqrt{n+1}.\left(\sqrt{n+1}+\sqrt{n}\right)}=\dfrac{1}{\sqrt{n}.\sqrt{n+1}}.\dfrac{1}{\sqrt{n+1}+\sqrt{n}}=\dfrac{1}{\sqrt{n}.\sqrt{n+1}}.\dfrac{\sqrt{n+1}-\sqrt{n}}{n+1-n}=\dfrac{1}{\sqrt{n}.\sqrt{n+1}}.\left(\sqrt{n+1}-\sqrt{n}\right)=\dfrac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n}.\sqrt{n+1}}\)
NM
14 tháng 6 2017
Áp dụng : \(\dfrac{1}{\sqrt{n}}>2\left(\sqrt{n+1}-\sqrt{n}\right)\)
\(\dfrac{1}{\sqrt{n}}+\dfrac{1}{\sqrt{n-1}}+...+\dfrac{1}{\sqrt{3}}+\dfrac{1}{\sqrt{2}}+1>2\left(\sqrt{n+1}-\sqrt{n}\right)+2\left(\sqrt{n}-\sqrt{n-1}\right)+...+2\left(\sqrt{4}-\sqrt{3}\right)+2\left(\sqrt{3}-\sqrt{2}\right)+2\left(\sqrt{2}-1\right).\)
\(=2\left(\sqrt{n+1}-1\right).\)
Ta có : \(\dfrac{1}{\left(n+1\right)\sqrt{n}}=\dfrac{\sqrt{n}}{n\left(n+1\right)}=\sqrt{n}\left(\dfrac{1}{n}-\dfrac{1}{n+1}\right)=\sqrt{n}\left(\dfrac{1}{\sqrt{n}}-\dfrac{1}{\sqrt{n+1}}\right)\left(\dfrac{1}{\sqrt{n}}+\dfrac{1}{\sqrt{n+1}}\right)=\left(1+\dfrac{\sqrt{n}}{\sqrt{n+1}}\right)\left(\dfrac{1}{\sqrt{n}}-\dfrac{1}{\sqrt{n+1}}\right)< 2\left(\dfrac{1}{\sqrt{n}}-\dfrac{1}{\sqrt{n+1}}\right)\)
Áp dụng điều này vào bài toán , ta có :
\(\left\{{}\begin{matrix}\dfrac{1}{2\sqrt{1}}< 2\left(1-\dfrac{1}{\sqrt{2}}\right)=2-\sqrt{2}\\\dfrac{1}{3\sqrt{2}}< 2\left(\dfrac{1}{\sqrt{2}}-\dfrac{1}{\sqrt{3}}\right)=\sqrt{2}-\dfrac{2}{\sqrt{3}}\\....\\\dfrac{1}{\left(n+1\right)\sqrt{n}}< 2\left(\dfrac{1}{\sqrt{n}}-\dfrac{1}{\sqrt{n+1}}\right)=\dfrac{2}{\sqrt{n}}-\dfrac{2}{\sqrt{n+1}}\end{matrix}\right.\)
\(\Rightarrow\dfrac{1}{2\sqrt{1}}+\dfrac{1}{2\sqrt{3}}+...+\dfrac{1}{\left(n+1\right)\sqrt{n}}< 2-\dfrac{2}{\sqrt{n+1}}< 2\) ( Sửa đề ^-^ )
cẩn thận khúc sau 1 chút