a(2n-1) = 1/2^n,
a(2n) = 1/[3*2^(n-1)].
s(2n) = a(1)+a(3)+...+a(2n-1) + a(2)+a(4)+...+a(2n) = (1/2)[1 + 1/2 + ... + 1/2^(n-1)] + (1/3)[1 + 1/2 + ... + 1/2^(n-1)] = (5/6)[ 1 + 1/2 + ... + 1/2^(n-1)] = (5/6)[1 - 1/2^n]/(1-1/2)]
= (5/3)[ 1 - 1/2^n]
= 5/3 - 5/[3*2^n]
s(2n-1) = s(2n) - a(2n) = 5/3 - (5/3)[1/2^n] - 1/[3*2^(n-1)]
= 5/3 - 5/[3*2^n] - 2/[3*2^n]
= 5/3 - 7/[3*2^n].
n->无穷大时, s(2n) = 5/3 - 5/[3*2^n] -> 5/3.
s(2n-1) = 5/3 - 7/[3*2^n] ->5/3.
因此,n->无穷大时,s(n) -> 5/3.
无穷数列1/2,1/3,1/4,1/6,……,1/2^n,1/3*2^n-1……的各项和为 5/3