Question

证明1+1/2^2+1/3^2+...+1/n^2<(2n-1)/n

用数学归纳法证明上述不等式(n>=2,n为正整数)

Answer 1

证明：1+1/2^2+1/3^2+...+1/n^2<(2n-1)/n (n>=2,n属于N*)
1）1+1/2^2=5/4 < 3/2
2) 设：1+1/2^2+1/3^2+...+1/k^2<(2k-1)/k,
1+1/2^2+1/3^2+...+1/k^2+1/(k+1)^2<(2k-1)/k+1/(k+1)^2
=(2k^3+4k^2+2k-k^2-2k-1+k)/k(k+1)^2
=2-(k-1)/k(k+1)<[2(k+1)-1]/(k+1)
也就是如果n=k时成立能推出n=k+1也成立
所以，1+1/2^2+1/3^2+...+1/n^2<(2n-1)/n

Answer 2

首先
1+1/2^2=5/4<(2n-1)/n=3/2

所以设1+1/2^2+1/3^2+...+1/k^2<(2k-1)/k成立
看k+1
1+1/2^2+1/3^2+...+1/k^2+1/（k+1）^2
<(2k-1)/k + 1/（k+1）^2
=[(2k-1)(k+1)+1]/（k+1）^2
=[2k^2+k]/（k+1）^2
<(2k+1)/k+1
=(2(k+1)-1)/(k+1)
成立
所以1+1/2^2+1/3^2+...+1/n^2<(2n-1)/n