a_n = ln(n^3)/(2n) Determine the convergence or divergence of the sequence with the given n'th term. If the sequence converges, find its limit.

a_n=(ln(n^3))/(2n)
The first few terms of the sequence are:
0 ,  0.5199 ,  0.5493 ,  0.5199 ,  0.4828 ,  0.4479 ,  0.4170 ,...
To determine if the sequence converge as the n becomes larger, take the limit of the nth-term as n approaches infinity.
lim_(n->oo)a_n
 =lim_(n->oo) (ln(n^3))/(2n)
To take the limit of this, apply  L'Hospital's Rule.
=lim_(n->oo) ((ln(n^3))')/((2n)')
=lim_(n->oo) (1/n^3*3n^2)/2
=lim_(n->oo) (3/n)/2
=lim_(n->oo) 3/(2n)
= 3/2 lim_(n->oo) 1/n
=3/2*0
=0
Therefore, the sequence is convergent.  And the terms converges to a value of 0.