Question

n为正整数,求证：4n/（4 n∧2+1）>ln（2n+1）/（2n-1）

Answer 1

很明显左右两式都大于零所以我们用左边除以右边在证明其小于1即可，左式除以右式得

(8*x^2)/((4*x^2+1)*ln(2*x+1))-(4*x)/((4*x^2+1)*ln(2*x+1))，求导得

((32*x^3+48*x^2+8*x-4)*ln(2*x+1)-64*x^4+32*x^3-16*x^2+8*x)/((32*x^5+16*x^4+16*x^3+8*x^2+2*x+1)*ln(2*x+1)^2)

开玩笑，根本求不出来

这时我们就要对上式进行简化，我们发现

4*x^2+1<4*x^2+1

所以4*x^2/(4*x^2+1)<4*x/(4*x^2)=1/x,因此要证明原式，只要证明

1/x<ln(2*x+1)/(2*x-1)

只需证(2*x-1)/x<ln(2*x+1)

再令f(x)=(2*x-1)/x-ln(2*x+1)，这时证明此时小于0就容易多了，剩余过程如下

f'(x)=-(2*x^2-2*x-1)/(2*x^3+x^2)

=-(2*x^2-2*x-1)/(x^2*(2*x+1))

当x>=sqrt(3)/2+1/2时,f'(x)<0

所以当x>sqrt(3)/2+1/2时，f(x)<f(x>=sqrt(3)/2+1/2)=0

所以当n>=2时(4*n^2)/(4*n^2+1)<1/n<ln(2*n+1)/(2*n-1)

当n=1时，明显4/5<ln3

原式的证

希望有帮助

Answer 2

设f(n)=4n/（4 n^2+1）-ln[（2n+1）/（2n-1）],则f'(n)=4(4n^2+1-8n^2)/(4n^2+1)^2-(2n-1)/(2n+1)*(-4)/(2n-1)^2=4(1-4n^2)/(4n^2+1)^2+4/(4n^2-1)=4[-(4n^2-1)^2+(4n^2+1)^2]/[(4n^2-1)(4n^2+1)^2]=64n^2/[(4n^2-1)...