Giúp mih với 

Đầ sai rồi đấy. Viết lại đi

Ta có: \(S=\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+...+\frac{1}{2019!}=1+\frac{1}{2!}+\frac{1}{3!}+...+\frac{1}{2019!}\)

Đặt \(M=\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+...+\frac{1}{2019!}\)

\(\Rightarrow M< \frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{2018\cdot2019}\)

\(\Rightarrow M< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2018}-\frac{1}{2019}\)

\(\Rightarrow M< 1-\frac{1}{2019}=\frac{2019}{2019}-\frac{1}{2019}=\frac{2018}{2019}\)

\(\Rightarrow S< 1+\frac{2018}{2019}=\frac{2019}{2019}+\frac{2018}{2019}=\frac{4037}{2019}< 2\)

\(\Rightarrow S< 2\) ( ĐPCM )

\(\frac{1}{2^2}< \frac{1}{1\cdot2}\\ \frac{1}{3^2}< \frac{1}{2\cdot3}\\ \frac{1}{4^2}< \frac{1}{3\cdot4}\\ ...\\ \frac{1}{n^2}< \frac{1}{\left(n-1\right)\cdot n}\)

\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{n^2}< \frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{\left(n-1\right)\cdot n}\\ \Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{n^2}< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{n-1}-\frac{1}{n}\\ \Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{n^2}< 1-\frac{1}{n}< 1\\ \Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{n^2}< 1\left(\text{với }n\in N;n\ge2\right)\)

A đâu !!

anh cũng đang định hỏi câu này

Ta có \(A=\frac{1}{2!}+\frac{2}{3!}+...+\frac{2014}{2015!}\)

=>  \(A=\frac{2-1}{2!}+\frac{3-1}{3!}+...+\frac{2015-1}{2015!}\)

=>  \(A=\frac{2}{2!}-\frac{1}{2!}+\frac{3}{3!}-\frac{1}{3!}+...+\frac{2015}{2015!}-\frac{1}{2015!}\)

=> \(A=1-\frac{1}{2!}+\frac{1}{2!}-\frac{1}{3!}+...+\frac{1}{2014!}-\frac{1}{2015!}\)

=>  \(A=1-\frac{1}{2015!}< 1\)

Mình đoán bạn thi học sinh giỏi. Bạn yên tâm đi, lớp 6 chưa hoc ! ( than cảm) đâu nên cô sẽ ko mắng. Mình cũng thi, cô bảo ko phải làm đó