Question

Cho n thuộc Z và n>1

Cm \(\frac{1}{n+1}\)+\(\frac{1}{n+2}\)+\(\frac{1}{n+3}\)+...+\(\frac{1}{n+n}\)>\(\frac{1}{2}\)

Answer 1

Ta có : 

\(\frac{1}{n+1}>\frac{1}{n+n}=\frac{1}{2n}\)

\(\frac{1}{n+2}>\frac{1}{n+n}=\frac{1}{2n}\)

\(\frac{1}{n+3}>\frac{1}{n+n}=\frac{1}{2n}\)

......................

\(\frac{1}{n+n}=\frac{1}{n+n}=\frac{1}{2n}\)

Cộng vế với vế ta được :

\(\frac{1}{n+1}+\frac{1}{n+2}+\frac{1}{n+3}+....+\frac{1}{n+n}>\frac{1}{2n}+\frac{1}{2n}+\frac{1}{2n}+....+\frac{1}{2n}\)( có n số \(\frac{1}{2n}\) )

\(\Rightarrow\frac{1}{n+1}+\frac{1}{n+2}+\frac{1}{n+3}+....+\frac{1}{n+n}>\frac{n}{2n}=\frac{1}{2}\) ( đpcm )