≤|〖〖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 )).