9th APMO 1997

Let T_n = 1 + 2 + ... + n = n(n+1)/2. Let S_n= 1/T₁ + 1/T₂ + ... + 1/T_n. Prove that 1/S₁ + 1/S₂ + ... + 1/S₁₉₉₆ > 1001.

Solution

1/T_m = 2(1/m - 1/(m+1) ). Hence S_n/2 = 1 - 1/(n+1). So 1/S_n = (1 + 1/n)/2. Hence 1/S₁ + 1/S₂ + ... + 1/S_n = 1996/2 + (1+ 1/2 + 1/3 + ... + 1/1996)/2.

Now 1 + 1/2 + (1/3 + 1/4) + (1/5 + 1/6 + 1/7 + 1/8) + (1/9 + ... + 1/16) + (1/17 + ... + 1/32) + (1/33 + ... + 1/64) + (1/65 + ... + 1/128) + (1/129 + ... + 1/256) + (1/257 + ... + 1/512) + (1/513 + ... + 1/1024) > 1 + 1/2 + 1/2 + ... + 1/2 = 6. So 1/S₁ + 1/S₂ + ... + 1/S_n = 1996/2 + 6/2 = 998 + 3 = 1001.

9th APMO 1997

© John Scholes
jscholes@kalva.demon.co.uk
11 Apr 2002