Nguyen svtkvtm Khôi Bùi Nguyễn Việt Lâm Lê Anh Duy Nguyễn Thành Trương DƯƠNG PHAN KHÁNH DƯƠNG An Võ (leo) Ribi Nkok Ngok Bonking ...

\(\frac{1}{2^2}< \frac{1}{1.2}\)

\(\frac{1}{3^2}< \frac{1}{2.3}\)

........

\(\frac{1}{n^2}< \frac{1}{\left(n-1\right)n}\)

=> \(A< \frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{n\left(n-1\right)}=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{n-1}-\frac{1}{n}=1-\frac{1}{n}< 1\)

Đpcm 

b)B=1/4(1/2^2+1/3^2+...+1/n^2)=1/4*A<1/4

Đặt P = ... 

* Chứng minh P > 1/2 : 

\(P\ge\frac{\left(1+1+1+...+1\right)^2}{n+1+n+2+n+3+...+n+n}\)

Từ \(n+1\) đến \(n+n\) có n số => tổng \(\left(n+1\right)+\left(n+2\right)+\left(n+3\right)+...+\left(n+n\right)\) là: 

\(\frac{n\left(n+n+n+1\right)}{2}=\frac{n\left(3n+1\right)}{2}\)

\(\Rightarrow\)\(P\ge\frac{n^2}{\frac{n\left(3n+1\right)}{2}}=\frac{2n}{3n+1}\)

Mà \(n>1\)\(\Leftrightarrow\)\(4n>3n+1\)\(\Leftrightarrow\)\(\frac{n}{3n+1}>\frac{1}{2}\)

\(\Rightarrow\)\(P>\frac{1}{2}\)

* Chứng minh P < 3/4 : 

Có: \(\frac{1}{n+1}\le\frac{1}{4}\left(\frac{1}{n}+1\right)\)

\(\frac{1}{n+2}\le\frac{1}{4}\left(\frac{1}{n}+\frac{1}{2}\right)\)

\(\frac{1}{n+3}\le\frac{1}{4}\left(\frac{1}{n}+\frac{1}{3}\right)\)

... 

\(\frac{1}{n+n}=\frac{1}{2n}=\frac{1}{4}\left(\frac{1}{n}+\frac{1}{n}\right)\)

\(\Rightarrow\)\(P\le\frac{1}{4}\left(\frac{1}{n}+1+\frac{1}{n}+\frac{1}{2}+\frac{1}{n}+\frac{1}{3}+...+\frac{1}{n}+\frac{1}{n}\right)\)

\(\Leftrightarrow\)\(P\le\frac{1}{4}\left(\frac{1}{n}+\frac{1}{n}+\frac{1}{n}+...+\frac{1}{n}\right)+\frac{1}{4}\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{n}\right)\)

\(\Leftrightarrow\)\(P\le\frac{1}{4}\left(n.\frac{1}{n}\right)+\frac{1}{4}\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{n}\right)< \frac{1}{4}+\frac{1}{4}=\frac{2}{4}< \frac{3}{4}\) ( do n>1 ) 

\(\Rightarrow\)\(P< \frac{3}{4}\)

Ta có : \(\frac{1}{\left(k+1\right)\sqrt{k}}=\frac{\sqrt{k}}{k\left(k+1\right)}=\sqrt{k}\left(\frac{1}{k\left(k+1\right)}\right)=\sqrt{k}\left(\frac{1}{k}-\frac{1}{k+1}\right)=\sqrt{k}\left(\frac{1}{\sqrt{k}}-\frac{1}{\sqrt{k+1}}\right)\left(\frac{1}{\sqrt{k}}+\frac{1}{\sqrt{k+}}\right)\)

\(=\left(1+\frac{\sqrt{k}}{\sqrt{k+1}}\right)\left(\frac{1}{\sqrt{k}}-\frac{1}{\sqrt{k+1}}\right)< 2\left(\frac{1}{\sqrt{k}}-\frac{1}{\sqrt{k+1}}\right)\)

Áp dụng : \(\frac{1}{2}+\frac{1}{3\sqrt{2}}+...+\frac{1}{\left(n+1\right)\sqrt{n}}< 2\left(\frac{1}{1}-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)=2\left(1-\frac{1}{\sqrt{n+1}}\right)=2-\frac{2}{\sqrt{n+}}< 2\)

Vậy ta có điều phải chứng minh.

Không có điều kiện gì à?Nếu n = 1 \(C>1>\frac{1}{4}\) vậy c/m làm gì?

Chắc là n>=3 đó