luam prima parte a execritiului:
[tex] x(1+ \frac{1}{2}+ \frac{1}{3}+ \frac{1}{3}+....+ \frac{1}{2014})[/tex]
o rescriem in felul urmator:
[tex]x[1+(1- \frac{1}{2} )+(1- \frac{2}{3} )+(1- \frac{3}{4} )+...+(1- \frac{2013}{2014} )][/tex]
daca aduma toti unu o sa avem:
[tex]x(2014- \frac{1}{2}- \frac{2}{3}- \frac{3}{4}-...- \frac{2013}{2014}) [/tex]
deci x =1
