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

A = 1/1.2.3 + 1/2.3.4 + ... + 1/18.19.20 

\(A=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{18.19}-\frac{1}{19.20}\right)\)

\(A=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{19.20}\right)\)

\(A=\frac{1}{4}-\frac{1}{2.19.20}< \frac{1}{4}\)

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.

Ta có: 1.2.3.4...2004 = 1.2.3.4.5...401...2004 = [5.401].1.2.3.4.6....2004 = 2005.1.2.3....2004 chia hết cho 2005

=> Khi nhân với 1 + 1/2 + ... + 1/2004 cũng chia hết cho 2005

AI THẤY ĐÚNG NHỚ ỦNG HỘ

Ta có: \(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2004}\)

\(=\left(1+\frac{1}{2004}\right)+\left(\frac{1}{2}+\frac{1}{2003}\right)+\left(\frac{1}{3}+\frac{1}{2002}\right)+...+\left(\frac{1}{1002}+\frac{1}{1003}\right)\)

\(=\frac{2005}{1.2004}+\frac{2005}{2.2003}+\frac{2005}{3.2002}+...+\frac{2005}{1002.1003}\)

\(=2005\left(\frac{1}{1.2004}+\frac{1}{2.2003}+\frac{1}{3.2002}+....+\frac{1}{1002.1003}\right)\)

\(\Rightarrow A=1.2.3.....2004.\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2004}\right)\)\(=1.2.3.....2004.2005\left(\frac{1}{1.2004}+\frac{1}{2.2003}+....+\frac{1}{1002.1003}\right)\)chia hết cho 2005 (đpcm)