Calculus of a Single Variable, Chapter 9, 9.2, Section 9.2, Problem 47

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.