x+1-1/2 +x+1-1/3 +...+x +1-1/4+....+x+1-1/n=n
nx + (1+1+...+1)- (1/2+1/3+....+1/n)=n
in rima paranteza avem n de 1
nx+n=n+(1/2+1/3+...1/n)
nx=1/2+1/3+...1/n
vom nota
1/2+1/3+...1/n cu SAn seria armonica n, pt care nu exista formula functie de n
deci x=SAn/n
sau
(∑ 1/k)/n unde k=2÷n
