Cho S=1/2+2/2^2+3/2^3+4/2^4+...+2003/2^2003

Ta có:
S = 1/2 + 2/2^2 + 3/2^3 + 4/2^4 + ... + 2003/2^2003
=> S = 1/2 + 2/4 + 3/8 + 4/16 + ... + 2003/2^2003
=> S = (1/2 + 1/4 + 1/8 + 1/16 + ... + 1/2^2003) + (1/4 + 1/8 + 1/16 + ... + 1/2^2003) + (1/8 + 1/16 + ... + 1/2^2003) + ... + (1/2^2003)
=> S = (1 - 1/2^2003) + (1/2 - 1/2^2002) + (1/4 - 1/2^2001) + ... + (1/2^2002 - 1/2)
=> S = 1 + 1/2 + 1/4 + ... + 1/2^2002 - 1/2^2003
=> S = 2 - 1/2^2003

Vì 0 < 1/2^2003 < 1 nên ta có:
1 < 2 - 1/2^2003 < 2
=> 1 < S < 2

Vậy ta đã chứng minh được 1 < S < 2.