≤|〖〖a_n〗_+〗_1|+|〖〖a_n〗_+〗_2|+...+|am|

　　≤1/2^(n+1) +1/2^(n+2) +...+1/2^m

　　=(1/2^(n+1) [1"-" (1/2)^(m"-" n) ])/(1"-" 1/2)

　　=1/2^n [1"-" (1/2)^(m"-" n) ]<1/2^n .

10.导学号26394040若数列{xn}的通项公式为xn=n/(n+1),求证x1·x3·x5·...·x2n-1<√((1"-" x_n)/(1+x_n )).

证明因为√((1"-" x_n)/(1+x_n ))=√((1"-" n/(n+1))/(1+n/(n+1)))=√(1/(2n+1)),

　　又((2n"-" 1)/2n)/√((2n"-" 1)/(2n+1))=(√(2n"-" 1) "·" √(2n+1))/2n

　　=√(4n^2 "-" 1)/2n<√(4n^2 )/2n=1,

　　所以(2n"-" 1)/2n<√((2n"-" 1)/(2n+1)),

　　所以x1·x3·x5·...·x2n-1

　　=1/2×3/4×...×(2n"-" 1)/2n

　　<√(1/3×3/5×"..." ×(2n"-" 1)/(2n+1))=√(1/(2n+1)),

　　故x1·x3·x5·...·x2n-1<√((1"-" x_n)/(1+x_n )).