Se aplica formula
1/(n+1)<ln(n+1)-ln n>1 =>
1/2<ln2-ln1
1./3<ln3-ln2
.............................
1/n<ln n-ln(n-1) se aduna termen cu termen
1/2+1/3+...+1/n<lnn-ln1 ln1=0
Se aduna 1 in ambii membrii
1+1/2+1/3+...+1/n<lnn+1=>
1+1/2+1/3+...+1/n-ln<1 in stanga ai obtinut sirul an
evident
1<an<1 =>an→1
