证明:1/100+1/101+...+1/119+1/120在2/11与1/5之间.

问题描述:

证明:1/100+1/101+...+1/119+1/120在2/11与1/5之间.
RT

1/(110-X)+1/(110+X)=220/[(110-X)*(110+X)]=220/(110*110-X*X)>220/(110*110)=2/110所以
1/100+1/101+...+1/119+1/120>10*2/110+1/110=21/110>20/110=2/11
同理1/100+1/120>1/101+1/119>...>1/108+1/112>1/109+1/111,所以
1/100+1/101+...+1/119+1/120