N=[ 1/ε]+ 1 is taken to ensure that the absolute value of the difference between the nth term and all subsequent terms and 2 (actually, 1/n) is less than ε.
So take N=[ 1/ε]+2, and N=[ 1/ε]+3. . . Both will do.
As for adding one more 1 instead of directly taking N=[ 1/ε], it is mainly to satisfy | an-2 |.
Actually, because [1/ε] < = 1/ε, actually1/[1/ε] > = 1/( 1/ε)=ε ,
After adding 1, n = [1/ε]+1> 1/ε, so 1/n