1+1/(2^2)+1/(3^2)+1/(4^2)+…+1/(n^2)<1-1/n 用数学归纳法证明

Question

n>=2,n∈N*

Answer 1

1.当n=2时，左边=1/2^2=1/4，右边=1-1/2=1/2，左边<右边，成立
2.假设当n=k时，1/2^2+1/3^2+...+1/k^2<1-1/k
所以：当n=k+1时，左边=1/2^2+1/3^2+...+1/k^2+1/(k+1)^2
<1-1/k+1/(k+1)^2
=[1-1/(k+1)]+[1/(k+1)-1/k+1/(k+1)^2]
=[1-1/(k+1)]+(k^2+k-k^2-2k-1+k)/k(k+1)^2
=[1-1/(k+1)]-1/k(k+1)^2
<1-1/(k+1)
得证。

Answer 2

注：从而n+1到3n,左边共有2n项．
（1）当n=2时,左=1/3 +1/4+1/5+1/6=57/60>54/60=9/10,成立．
（2）假设n=k时,有1/(k+1) +1/(k+2) +...+1/3k >9/10
那么　1/(k+2)+1/(k+3) +...+1/3(k+1)
=[1/(k+1) +1/(k+2)+...+1/3k] +1/(3k+1) +1/(3k+2)+1/(3k+3) -1/(k+1)
>9/10 +1/(3k+3) +1/(3k+3)+1/(3k+3) -1/(k+1)
=9/10
即n=k+1时命题也成立,
从而　原不等式对n∈N,且n>1成立．