Re:RES: [obm-l] desigualdade

Ou então, P = (1/2)*(3/4)*(5/6)*...*(99/100) Q = (2/3)*(4/5)*(6/7)*...*(100/101) Claramente, P Q == P^2 PQ = 1/101 == P 1/raiz(101) 1/raiz(100) = 1/10 Por outro lado, R = (1/2)*(2/3)*(4/5)*...* (98/99), de modo que: P R == P^2 PR = (1/2)*(1/100) = 1/200 == P 1/raiz(200) 1/raiz(225) =

Re:RES: [obm-l] desigualdade

Eh, nao ha incoerencia nenhuma, pois 1/15 =0,0666... 0,096849. Eu fiz conta errada Artur --- claudio.buffara [EMAIL PROTECTED] wrote: Ou então, P = (1/2)*(3/4)*(5/6)*...*(99/100) Q = (2/3)*(4/5)*(6/7)*...*(100/101) Claramente, P Q == P^2 PQ = 1/101 == P 1/raiz(101)