已知一冪函數(shù)與它的反函數(shù)為同一函數(shù),且y=f(x)在（0,∞）上單調(diào)遞減,
記S(n)=f(1)+(2)+f(3)+……+f(n) (n∈N)
(1)、分別求出使下列兩個(gè)不等式各自成立的最小自然數(shù)m及k：S(m)﹥1+S(1)；S(k)﹥1+S(2)；
(2)、求最小自然數(shù)a ,使對一切n∈N,不等式S(3n+a)﹥1+S(n)成立.
設(shè)此冪函數(shù)為y=x^μ；其反函數(shù)為y=x^(1/μ)；二者為同一函數(shù),且在（0,∞）上單調(diào)遞減,
故μ=-1；即此冪函數(shù)為y=x⁻¹=1/x；
(1).S(m)=1+(1/2)+(1/3)+.+(1/m)>0.577216.+lnm=1+S(1)=1+1=2
故得lnm=2-0.577216.=1.422784...,；由此得m=4；即1+(1/2)+(1/3)+(1/4)=25/12=22.0833...>2
S(k)=1+(1/2)+.+(1/k)>0.577216.+lnk=1+S(2)=1+1+(1/2)=2.5
故得lnk=2.5-0.577216...=1.922784.；故k=7；即1+(1/2)+.+(1/7)=1090/420=2.595>2.5.
(2).S(3n+a)=1+(1/2)+...+(1/n)+1/(n+1)+1/(n+2)+.+(1/2n)+1/(2n+1)+1/(2n+2)+.+(1/3n)+
+1/(3n+1)+1/(3n+2)+.+1/(3n+a)>1+S(n)=1+1+(1/2)+(1/3)+.(1/n)
即有1/(n+1)+1/(n+2)+.+(1/2n)+1/(2n+1)+...+1/(2n+2)+.+(1/3n)+1/(3n+1)+1/(3n+2)+
.+1/(3n+a)>1；當(dāng)n=1時(shí)有(1/2)+(1/2)+(1/3)+(1/4)=1+(7/12)>1,故mina=1.