`sum_(n=1)^oo 3^n/n^3`

To determine if the series is convergent or divergent, apply the ratio test. The formula for the ratio test is:

`L = lim_(n->oo) |a_(n+1)/a_n|`

If L<1, the series converges.

If L>1, the series diverges.

And if L=1, the test is inconclusive.

Applying the formula above, the value of L will be:

`L = lim_(n->oo) | (3^(n+1)/(n+1)^3)/(3^n/n^3)|`

`L= lim_(n->oo) | 3^(n+1)/(n+1)^3 * n^3/3^n|`

`L = lim_(n->oo) | (3n^3)/ (n+1)^3|`

`L= lim_(n->oo) | (3n^3)/(n^3+3n^2+3n+1)|`

`L=lim_(n->oo) |(3n^3)/(n^3(1+3/n+3/n^2+1/n^3))|`

`L= lim_(n->oo) |3/(1+3/n+3/n^2+1/n^3)|`

`L= 3/(1+0+0+0)`

`L=3`

**Therefore, the series diverges.**