1^s +1/2^s +1/3^s +1/4^s + ...

maximum s=1

2011-01-16 00:17:36 補充：
1. 1/k > 1/x , for x in [k, k+1]
thus, Σ[k=1~∞] 1/k > ∫[1~∞] 1/x dx = ∞
hence, (s=1) 1/1+ 1/2+ 1/3+... div.
2. if s<1 , 1/1^s+ 1/2^s + 1/3^s+ ... > 1/1+ 1/2+ 1/3+ ...
hence, when s<=1, 1/1^s+ 1/2^s+ 1/3^s+ ... div.
3. in the case of s>1
1/1^s + 1/2^s + 1/3^s + ...
< 1+ ∫[1~∞] 1/x^s dx = 1 + 1/(s-1)
and 1/k^s >0
then 1/1^s+ 1/2^s + 1/3^s + ... conv.
so, 1/1^s + 1/2^s + 1/3^s + ... div. then s<= 1
the maximum of s is 1.