Nhận thấy 1/1.2.3 = 1/2.3;    1/1.2.3.4 < 1/3.4;   1/1.2.3.4.5 < 1/4.5;                               1/1.2.3...n  < 1/n(n-1)

=> 1 + 1/1.2 + 1/1.2.3 +... + 1/1.2.3...n < 1 + 1/1.2 + 1/2.3 + 1/3.4 + ... + 1/n(n-1)

=>  1 + 1/1.2 + 1/1.2.3 +... + 1/1.2.3...n < 1 + 1 -1/2 + 1/2 - 1/3 + 1/3 - 1/4 +...+            1/n-1 - 1/n

=>1 + 1/1.2 + 1/1.2.3 +... + 1/1.2.3...n < 2  - 1/n < 2

=> đpcm

Lời giải:

\(A=\frac{1}{1.2}+\frac{2}{1.2.3}+\frac{3}{1.2.3.4}+...+\frac{2011}{1.2.3...2012}\)

\(=\frac{2-1}{1.2}+\frac{3-1}{1.2.3}+\frac{4-1}{1.2.3.4}+...+\frac{2012-1}{1.2.3...2012}\)

\(=1-\frac{1}{1.2}+\frac{1}{1.2}-\frac{1}{1.2.3}+\frac{1}{1.2.3}-\frac{1}{1.2.3.4}+...+\frac{1}{1.2.3...2011}-\frac{1}{1.2.3...2012}\)

\(=1-\frac{1}{1.2...2012}< 1\)

Ta có đpcm.

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)