bạn ơi hình như đề bài là: 

\(A=\frac{1}{1+2}+\frac{1}{1+2+3}+...+\frac{1}{1+2+3+...+100}\)thì phải ha.

Đặt A = \(1+\frac{1}{1.2}+\frac{1}{1.2.3}+\frac{1}{1.2.3.4}+...+\frac{1}{1.2.3....n}\)

Ta có: \(\frac{1}{1.2}=\frac{1}{1.2}\)

\(\frac{1}{1.2.3}=\frac{1}{2.3}\)

\(\frac{1}{1.2.3.4}< \frac{1}{3.4}\)

..............

\(\frac{1}{1.2.3....n}< \frac{1}{\left(n-1\right)n}\)

Cộng vế với vế ta được:

\(A< 1+\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{\left(n-1\right)n}=1+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{n-1}-\frac{1}{n}=1+1-\frac{1}{n}=2-\frac{1}{n}< 2\)(đpcm)

tính tổng dãy số thì dễ nhưng hãy viết rõ ràng hơn

bạn làm theo công thức \(\frac{n}{n.\left(n+1\right)}=\frac{n}{n}-\frac{n}{n+1}\)

a)Đặt A= \(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{\left(2n-1\right)\left(2n+1\right)}\)

\(\Rightarrow2A=\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{\left(2n-1\right)\left(2n+1\right)}\)

\(\Rightarrow2A=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{2n-1}-\frac{1}{2n+1}\)

\(\Rightarrow2A=1-\frac{1}{2n+1}< 1\)

\(\Rightarrow A< \frac{1}{2}\)(đpcm)

b)Ta có: \(1+\frac{1}{1.2}+\frac{1}{1.2.3}+\frac{1}{1.2.3.4}+...+\frac{1}{1.2.3...n}< 1+\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{\left(n-1\right)n}\)

mà \(1+\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{\left(n-1\right)n}=1+\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{n-1}-\frac{1}{n}\)

\(=1+1-\frac{1}{n}\)

\(=2-\frac{1}{n}< 2\)

\(\Rightarrow1+\frac{1}{1.2}+\frac{1}{1.2.3}+\frac{1}{1.2.3.4}+...+\frac{1}{1.2.3...n}< 1+\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{\left(n-1\right)n}< 2\)

\(\Rightarrow1+\frac{1}{1.2}+\frac{1}{1.2.3}+\frac{1}{1.2.3.4}+...+\frac{1}{1.2.3...n}< 2\)(đpcm)