Ta có: \(\frac{1}{2^2}=\frac{1}{2.2}< \frac{1}{1.2}\)

Tương tự : \(\frac{1}{3^2}< \frac{1}{2.3}\); \(\frac{1}{4^2}< \frac{1}{3.4}\); ......... ; \(\frac{1}{2014^2}< \frac{1}{2013.2014}\)

\(\Rightarrow S< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+........+\frac{1}{2013.2014}\)               

        \(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+.........+\frac{1}{2013}-\frac{1}{2014}\)

        \(=1-\frac{1}{2014}=\frac{2013}{2014}\)

\(\Rightarrow S< \frac{2013}{2014}\left(đpcm\right)\)

em thử nhân S với 5 rồi lấy 5S= S thử đi

chị làm toàn như vậy

ko bt có đc ko nữa

Bài 2)

Ta có \(\frac{a}{b}< \frac{c}{d}\)

\(\Rightarrow ad< bc\)

Xét \(\frac{a}{b}< \frac{a+c}{b+d}\)

\(\Rightarrow a\left(b+d\right)< b\left(a+c\right)\)

\(\Rightarrow ab+ad< ab+bc\)

\(\Rightarrow ad< bc\) ( thỏa mãn đề bài )

Vậy \(\frac{a}{b}< \frac{a+c}{b+d}\) (1)

Xét \(\frac{a+c}{b+d}< \frac{c}{d}\)

\(\Rightarrow d\left(a+c\right)< c\left(b+d\right)\)

\(\Rightarrow ad+cd< bc+cd\)

\(\Rightarrow ad< bc\) ( thỏa mãn đề bài )

Vậy \(\frac{a+c}{b+d}< \frac{c}{d}\) (2)

Từ (1) và (2)

\(\Rightarrow\frac{a}{b}< \frac{a+c}{b+d}< \frac{c}{d}\) (đpcm)

Đặt \(S=\frac{1}{2^2}+\frac{1}{3^2}+........+\frac{1}{2014^2}\)

Đặt A=\(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+........+\frac{1}{2014.2015}\)

\(A=\left(\frac{1}{2}-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{4}\right)+.....\left(\frac{1}{2014}-\frac{1}{2015}\right)\\ =>A=\frac{1}{2}-\frac{1}{2015}\\ =>A=\frac{2013}{4030}\)

Mà S>A =>S>\(\frac{2013}{4030}\)

gap A len 1/2

\(2A=1+\frac{1}{2}+\left(\frac{1}{2}\right)^2+...+\left(\frac{1}{2}\right)^{2015}\)

\(\Rightarrow2A-A=1-\left(\frac{1}{2}\right)^{2014}\Rightarrow A=1-\left(\frac{1}{2}\right)^{2014}< 1\)