Question

[Nguyễn Bá Phi]

giup với ạ:)))

Answer 1

[Nguyễn Lê Phước Thịnh]

a: \(\dfrac{1}{\sqrt{n}}=\dfrac{2}{\sqrt{n}+\sqrt{n}}< \dfrac{2}{\sqrt{n}+\sqrt{n-1}}\)

\(=2\left(\sqrt{n}-\sqrt{n-1}\right)\)

\(\dfrac{1}{\sqrt{n}}=\dfrac{2}{\sqrt{n}+\sqrt{n}}>\dfrac{2}{\sqrt{n}+\sqrt{n+1}}=2\left(\sqrt{n+1}-\sqrt{n}\right)\)

=>\(2\left(\sqrt{n+1}-\sqrt{n}\right)< \dfrac{1}{\sqrt{n}}< 2\left(\sqrt{n}-\sqrt{n-1}\right)\)

b: 2(căn 101-căn 2)<1/căn 2+1/căn 3+...+1/căn 100<2(căn 100-căn 1)=18

2(căn 101-căn 2)=2căn 101-2căn 2>2căn 100-3=17

=>\(17< \dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{3}}+...+\dfrac{1}{\sqrt{100}}< 18\)