`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.

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