11且n为整数） 不要用数学归纳法证明

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1/n+1/(n+1)+1/(n+2) +……+1/n^2>1/(n+1)+1/(n+1)+1/(n+2) +……+1/n^2>2/(n+1)+1/(n+2) +……+1/n^2>2/(n+2)+1/(n+2) +……+1/n^2>3/(n+2)+1/(n+3)……+1/n^2>``````
n^2+1/n^2>1======以下答案可供参考======
供参考答案1:
1/(n+1) > 1/n^21/(n+2) > 1/n^2....1/(n^2-1) > 1/n^21/n^2 = 1/n^2 相加，共 n^2 -n 项，因n>1, n^2-n >=2,所以1/(n+1)+1/(n+2) +……+1/n^2 > (n^2-n)/n^2 =1- 1/n
把-1/n移项到左边，得到：
1/n+1/(n+1)+1/(n+2) +……+1/n^2
> 1 得证。供参考答案2:
1/n+1/(n+1)+1/(n+2) +……+1/n^2
=1/n+[1/(n+1)+1/(n+2)+…+1/(2n)]+……+{1/[n(n-1)+1]+1/[n(n-1)+2]…+1/[n^2]}>1/n+1/2+1/3+…+1/n≥(1/n)*n=1 （n>1）供参考答案3:
很简单...
左边有n^2-n+1项,所以,将1平均分成n^2-n+1项,每项都减去1/(n^2-n+1)
显然1/n>1/(n+1) 1/(n+1)>1/(n+2)......
因此,左式中最小项为末项,
因此,只要比较1/(n^2)与1/(n^2-n+1)即可
显然,n>1时 1/(n^2)>1/(n^2-n+1)
因此,左式中每项减去1/(n^2-n+1)后仍大于0,所以原式证毕
不完全是归纳法....但肯定用到无穷推导...
供参考答案4:
1/n+1/(n+1)+1/(n+2) +……+1/n^2
>1/n+1/n^2+1/n^2 +……+1/n^2(除第1项外每项用较小项1/n^2代替)
=1/n+(n^2-n)(1/n^2)(n^2-n相同项相加)
=1/n+(n^2)(1/n^2)-n(1/n^2)=1