已知函数f（x）=x-ln(x+1),求证2+2/3+2/5+······2/2n-1-ln(2n+1)

Question

已知函数f（x）=x-ln(x+1),求证2+2/3+2/5+······2/2n-1-ln(2n+1)

Answer 1

数学之美团为你解答
令 g(x) = x - ln(x+1) - 1/2 x²
g'(x) = 1 - 1/(x+1) - x = - x² / (x+1) < 0
g(x)为减函数
当 x>0时,g(x) < g(0) = 0
即 x - ln(x+1) < 1/2 x²
分别令 x = 2 ,2/3 ,2/5 ,……
2 - ln3 = 2 - ln3 （第一项不能放缩,放缩后右边是2,再加其他项就超过2了）
2/3 - ln(5/3) < 1/2 *(2/3)²
2/5 - ln(7/5) < 1/2 *(2/5)²
……
2/(2n-1) - ln[(2n+1)/(2n-1)] < 1/2 *[2/(2n-1)]²
各式相加 2+2/3+2/5+……+2/(2n-1) - ln(2n+1) < 2 - ln3+ 2/3²+2/5²+2/7²+……+2/(2n-1)²
< 2 - ln3 + (1 - 1/3) + (1/3 - 1/5) + (1/5 - 1/7)+……+[ 1/(2n-1) - 1/(2n-3)]
= 2 - ln3 + 1 - 1/(2n-3) < 2 + (1 - ln3) < 2
证法二：
f'(x) = 1 - 1/(x+1) = x/(x+1)
当x>0时,f'(x)>0,f(x)单调递增
f(x)>f(0) = 0 即 x > ln(x+1)
令 x = - 1/n
- 1/n > ln(1 - 1/n) = ln [(n-1)/n] = ln(n-1) - lnn
1/n < lnn - ln(n-1)
2+2/3+2/5+……+2/(2n-1)
< 2 + (1/2+1/3)+(1/4+1/5)+……+[1/(2n-2)+1/(2n-1)]
< 2+ ln2-ln1+ln3-ln2+ln4-ln3+……+ln(2n-1) - ln(2n-2)
= 2+ln(2n-1)
< 2+ln(2n+1)
即 2+2/3+2/5+······2/(2n-1) - ln(2n+1) < 2