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已知函数f(x)=lnx+a/(x+1)(a属于R),求证ln(n+1)>1/3+1/5+1/7+...+1/(2n +

已知函数f(x)=lnx+a/(x+1)(a属于R),求证ln(n+1)>1/3+1/5+1/7+...+1/(2n +1)
万般无奈年孤独 1年前 已收到1个回答 我来回答 举报

hengyouqian 幼苗

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证明:考虑函数f(x)=ln(1+1/x)-1/(2x+1),x>0.显然当x->+∞时,f(x)=0.
而f'(x)=-1/[n*(n+1)]+2/[(2n+1)^2]=1/(2n^2+2n+1/2)-1/(n^2+n)=-(n^2+n+1/2)/[(2n^2+2n+1/2)*(n^2+n)]=-[(n+1/2)^2+1/4]/[(2n^2+2n+1/2)*(n^2+n)]0时为单调递减函数,则必有x>0时f(x)=ln(1+1/x)-1/(2x+1)>0,于是有ln(1+1/x)>1/(2x+1),也即当x>0时,有
ln(x+1)-lnx>1/(2x+1)成立.于是
ln2-ln1>1/3
ln3-ln2>1/5
ln4-ln3>1/7
……
lnn-ln(n-1)>1/(2n-1)
ln(n+1)-lnn>1/(2n+1)
前述不等式左右两边分别相加,便得
ln(n+1)>1/3+1/5+1/7+…+1/(2n +1)

1年前

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