Ta có:
1/5<1/4 ; 1/6<1/4 ; ... ; 1/8<1/4
⇒ 1/5+1/6+1/7+1/8<1/4 x 4
⇒ 1/5+1/6+1/7+1/8<1 (1)
Ta lại có:
1/9<1/7 ; 1/10<1/7 ; ... ; 1/17<1/7
⇒ 1/9+1/10+...+1/17<1/7 x 7
⇒ 1/9+1/10+...+1/17<1 (2)
Từ (1) và (2) suy ra:
1/5+1/6+1/7+.....+1/17=(1/5+1/6+1/7+1/8)+(1/9+1/10+...+1/17)
⇒1/5+1/6+1/7+.....+1/17< 1+1
⇒1/5+1/6+1/7+.....+1/17< 2
⇒D<2