Question

用数学归纳法证明不等式1/4+1/9+1/16+...+1/n²

Answer 1

当n=2时,不等式左边=1/4,右边=1-1/2=1/2；1/4＜1/2,不等式成立
假设当n=k(k≥2,k∈N*)时,原不等式成立,
即1/4+1/9+1/16+······+1/k²＜1-1/k
则当n=k+1时,
1/4+1/9+1/16+······+1/k²+1/(k+1)²＜1-1/k+1/(k+1)²=1-[1/k-1/(k+1)²]
现在比较1/k-1/(k+1)²和1/(k+1)的大小
1/k-1/(k+1)²-1/(k+1)=[(k+1)²-k-k(k+1)]/[k(k+1)²]=1/[k(k+1)²]＞0
所以1/k-1/(k+1)²＞1/(k+1)
则1-[1/k-1/(k+1)²]＜1-1/(k+1)
于是有1/4+1/9+1/16+······+1/k²+1/(k+1)²＜1-1/(k+1)
即当n=k+1时,原不等式亦成立
综上所述,不等式1/4+1/9+1/16+...+1/n²