Question

Aratati ca $\frac {1}{p+1}+ \frac {1}{p+2}+...+ \frac {1}{2p}> \frac {1}{2}$, oricare ar fi p>1.

Answer 1

   
$\frac {1}{p+1}+ \frac {1}{p+2}+...+ \frac {1}{2p} \\ \text{Sirul are p termeni.} \\\text{Luam urmatorul sir care are tot p termeni:} \\ \frac {1}{p+p}+ \frac {1}{p+p}+...+ \frac {1}{2p} = \frac {1}{2p}+ \frac {1}{2p}+...+ \frac {1}{2p}= \frac{p}{2p}= \frac{1}{2} \\ \\ Dar: \\ \\ \frac{1}{p+1}> \frac{1}{p+p} \\ \\ \frac{1}{p+2}> \frac{1}{p+p} \\ \\ samd \\ \\ =>\frac {1}{p+1}+ \frac {1}{p+2}+...+ \frac {1}{2p}> \frac {1}{2}$