求证:ln(n+1)>1/3+1/5+1/7+·······+1/(2n+1) (n∈N)

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求证:ln(n+1)>1/3+1/5+1/7+·······+1/(2n+1)(n∈N)求证:ln(n+1)>1/3+1/5+1/7+·······+1/(2n+1)(n∈N)求证:ln(n+1)>1

求证:ln(n+1)>1/3+1/5+1/7+·······+1/(2n+1) (n∈N)
求证:ln(n+1)>1/3+1/5+1/7+·······+1/(2n+1) (n∈N)

求证:ln(n+1)>1/3+1/5+1/7+·······+1/(2n+1) (n∈N)
因为1/(2x+1)是凹函数,所以1/3+1/5+...1/(2n+1)<积分(0到n)dx/(2x+1)=ln(2n+1)/2
因为lnx是凸函数,所以[ln1+ln(2n+1)]/2