求证1+1/3^2+1/5^2+..+1/(2n-1)^2>7/6-1/2(2n+1)(n>1)求证1+1/3^2+1/5^2+..+1/(2n-1)^2>7/6-1/2(2n+1)(n>1,n属于自然数)

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求证1+1/3^2+1/5^2+..+1/(2n-1)^2>7/6-1/2(2n+1)(n>1)求证1+1/3^2+1/5^2+..+1/(2n-1)^2>7/6-1/2(2n+1)(n>1,n属于自然数)

求证1+1/3^2+1/5^2+..+1/(2n-1)^2>7/6-1/2(2n+1)(n>1)求证1+1/3^2+1/5^2+..+1/(2n-1)^2>7/6-1/2(2n+1)(n>1,n属于自然数)
求证1+1/3^2+1/5^2+..+1/(2n-1)^2>7/6-1/2(2n+1)(n>1)
求证1+1/3^2+1/5^2+..+1/(2n-1)^2>7/6-1/2(2n+1)(n>1,n属于自然数)

求证1+1/3^2+1/5^2+..+1/(2n-1)^2>7/6-1/2(2n+1)(n>1)求证1+1/3^2+1/5^2+..+1/(2n-1)^2>7/6-1/2(2n+1)(n>1,n属于自然数)
对于n≥5,即2n-1≥9,有:
1+1/3^2+1/5^2+…+1/(2n-1)^2
=(1+1/3^2+1/5^2+1/7^2)+1/9^2+1/11^2+…+1/(2n-1)^2,
经计算1+1/3^2+1/5^2+1/7^2=1+1/9+1/25+1/49>1.17,
因此,
1+1/3^2+1/5^2+…+1/(2n-1)^2
=(1+1/3^2+1/5^2+1/7^2)+1/9^2+1/11^2+…+1/(2n-1)^2
>1.17+1/9^2+1/11^2+…+1/(2n-1)^2
>1.17+1/(9*11)+1/(11*13)+…+1/[(2n-1)(2n+1)]
=1.17+1/2*[(1/9-1/11)+(1/11-1/13)+…+1/(2n-1)-1/(2n+1)]
=1.17+1/2*[1/9-1/(2n+1)]
>1.17-1/[2(2n+1)]
由于7/61.17-1/[2(2n+1)]>7/6-1/[2(2n+1)]..
下面验证,123/21=7/6-1/14,成立,
n=4时,1+1/9+1/25+1/49=12916/11025>10/9=7/6-1/18,成立..
因此,对于任意的n>1,不等式成立..